Electrostatic Potential And Capacitance - Class 12 Medical Physics - Extra Questions
Three capacitors each of capacitance $$9\ pF$$ are connected in series as shown in figure.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor, if the combination is connected to a $$120\ volt$$ supply?
Three point charges $$Q_1=25\mu$$C, $$Q_2=50\mu$$C and $$Q_3=100\mu$$C, are kept at the corners A, B and C respectively of an equilateral triangle ABC having each side equal to $$7.5$$m. Calculate the total electrostatic potential energy of the system.
Explain series combination of Capacitors. Derive the formula for equivalent capacitance.
When one terminal of a capacitor is connected to the terminal of another capacitors , called series combination of capacitors. In series, each capacitor has same charge flow from battery. The three capacitors $$C_1, C_2$$ and $$C_3$$ are in series.
Now, charge on first capacitor is $$Q_1=C_1V_1$$, charge on second capacitor is $$Q_2=C_2V_2$$ and charge on third capacitor is $$Q_3=C_3V_3$$.
In series, $$Q_1=Q_2=Q_2=Q$$ and $$Q=C_{eq}V$$
Also, $$V=V_1+V_2+V_3$$
$$\dfrac{Q}{C_{eq}}=\dfrac{Q}{C_1}+\dfrac{Q}{C_2}+\dfrac{Q}{C_3}$$
$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$$, this is the formula for equivalent capacitance in series.
In the arrangement shown in fig. plate B has a charge equal to 60 $$\mu C$$ The ratio $$d_1/d_2$$ isThen
$$q_1=$$____
$$q_2=$$____
$$q_3=$$____
$$q_4=$$____
$$q_5=$$____
$$q_6=$$____
$${ d }_{ 1 }=2{ d }_{ 2 }$$
$${ q }_{ 3 }+{ q }_{ 4 }=60\mu c$$
$${ q }_{ 2 }=-{ q }_{ 3 }$$
$${ q }_{ 5 }=-{ q }_{ 4 }$$
$$\dfrac { { q }_{ 3 }\times { d }_{ 1 } }{ A{ \epsilon}_{ 0 } } =\dfrac { { q }_{ 4 }{ d }_{ 2 } }{ A{ \epsilon }_{ 0 } } $$
$${ q }_{ 3 }{ d }_{ 1 }={ q }_{ 4 }{ d }_{ 2 }$$
$$2{ q }_{ 3 }={ q }_{ 4 }$$
$$3{ q }_{ 3 }=60$$
$$\boxed { { q }_{ 3 }=20 } $$
so,
$${ q }_{ 1 }=20\mu c$$
$${ q }_{ 2 }=-20\mu c$$
$${ q }_{ 3 }=20\mu c$$
$${ q }_{ 4 }=40\mu c$$
$${ q }_{ 5 }=-40\mu c$$
$${ q }_{ 6 }=40\mu c$$
Two capacitors are connected as shown in figure below. If the equivalent capacitance of the combination is $$4 \mu F$$
(i) Calculate the value of $$C$$.
(ii) Calculate the charge on each capacitor.
(iii) What will be the potential drop across each capacitor?
Show that electric potential at a point P, at a distance 'r' from a fixed point charge Q, is given by $$\displaystyle V=\left(\displaystyle\frac{1}{4\pi \epsilon_0}\right)\displaystyle\frac{Q}{r}$$.
A parallel plate capacitor has circular plates of $$10\ cm$$ radius separated by an air gap of $$1\ mm$$. It is charged by connecting the plates to a $$100\ volt$$ battery. Find the change in energy stored in the capacitor when the plates are moved to a distance of $$1\ cm$$ and the plates are maintained in connection with the battery.
A parallel plate capacitor with air has a capacity of $$8\ pF$$. Calculate the capacity of the capacitor if the distance between the plates in halved and the space between them is fully filled with a substance of dielectric constant $$5$$.
Define electric potential and write down its dimension.
(a) A charged rod P attracts rod R whereas P repels another charged rod Q. What type of force is developed between Q and R?
(b) Define the capacitance of capacitor. Derive and expression for the capacity at a parallel plate capacitor with a dielectric slab placed in between the plates of capacitor.
For a capacitor, the distance between two plates is $$5x$$, the electric field between them is $$E_{0}$$, now dielectric slab having dielectric constant $$3$$ and thickness $$3x$$ is placed between them in contact with one plate. In this condition what is the potential difference between its two plates?
A network of four capacitors of capacity equal to $${C_1} = C,\,\,{C_2} = 2C,\,\,{C_3} = 3C$$ and $${C_4} = 4C$$ are connected to a battery as shown in the figure. The ratio of the charges on $${C_2}$$ and $${C_4}$$ is:
On what factors does the capacitance of a parallel plate capacitors depends?
Define electric potential due to a point charge and arrive at the expression for the electric potential at a point due to a point charge in its vicinity.
The electrostatic potential at any point in an electrostatic field is defined as the work done in carrying a unit positive charge from infinity to that point against the electrostatic force of the field.
Consider a point Charge $$+Q$$ at $$O$$.
Let $$P$$ be a point at a distance $$r$$ from $$O$$.
Consider a point $$A$$ at a distance $$x$$ from $$O$$.
The magnitude of electrostatic force on unit-positive charge at $$A$$, is
$$\quad F=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q\times i }{ { x }^{ 2 } } $$
$$F=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } $$ along $$OP$$
Work Done in moving a unit positive charge from $$A$$ to $$B$$ through a small distance $$dx$$ is,
$$\Delta W=\vec { F } .d\vec { x } =Fdx\cos { { 180 }^{ o } } $$
$$\Delta W=-Fdx$$
$$\Delta W=\cfrac { -Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } dx$$
Total work done in moving a unit positive change from infinity to the point $$P$$ is,
$$W=\displaystyle \int _{ \infty }^{ r }{ \cfrac { -1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q }{ { x }^{ 2 } } } dx=-\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } \int _{ \infty }^{ r }{ \cfrac { 1 }{ x^2 } } dx$$
$$W=-\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 } } { \left[ \cfrac { 1 }{ x } \right] }_{ \infty }^{ r }=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 } } \left[ \cfrac { 1 }{ r } -\cfrac { 1 }{ \infty } \right] $$
$$W=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }r }$$
By definition, this work done is equal to the electrostatic potential (V) at that point due to $$Q$$.
$$V=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q }{ r } $$
Consider a point Charge $$+Q$$ at $$O$$.
Let $$P$$ be a point at a distance $$r$$ from $$O$$.
Consider a point $$A$$ at a distance $$x$$ from $$O$$.
The magnitude of electrostatic force on unit-positive charge at $$A$$, is
$$\quad F=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q\times i }{ { x }^{ 2 } } $$
$$F=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } $$ along $$OP$$
Work Done in moving a unit positive charge from $$A$$ to $$B$$ through a small distance $$dx$$ is,
$$\Delta W=\vec { F } .d\vec { x } =Fdx\cos { { 180 }^{ o } } $$
$$\Delta W=-Fdx$$
$$\Delta W=\cfrac { -Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } dx$$
Total work done in moving a unit positive change from infinity to the point $$P$$ is,
$$W=\displaystyle \int _{ \infty }^{ r }{ \cfrac { -1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q }{ { x }^{ 2 } } } dx=-\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } \int _{ \infty }^{ r }{ \cfrac { 1 }{ x^2 } } dx$$
$$W=-\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 } } { \left[ \cfrac { 1 }{ x } \right] }_{ \infty }^{ r }=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 } } \left[ \cfrac { 1 }{ r } -\cfrac { 1 }{ \infty } \right] $$
$$W=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }r }$$
By definition, this work done is equal to the electrostatic potential (V) at that point due to $$Q$$.
$$V=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q }{ r } $$
In the following circuit potential at point 'A' is zero then determine Potential at each point.
Three capacitors having capacitances $$20\mu F, 30\mu F$$ and $$40\mu F$$ are connected in series with a $$12\ V$$ battery. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors?
Match the columns.
What is the electric potential at the center of the triangle in figure.
$$\because$$ Electric potential is a sealer quantity
$$\therefore$$ It can be added algebraically.
$$\Rightarrow V_{net}=V_{1}+V_{2}+V_{3}$$
$$\Rightarrow V_{net}=\dfrac{kq_{1}}{r}+\dfrac{kq_{2}}{r}+\dfrac{kq_{3}}{r}$$
$$\Rightarrow V_{net}=\dfrac{k}{r}(q_{1}+q_{2}+q_{3})$$
$$\Rightarrow V_{net}=\dfrac{9\times 10^{9}}{\left(\dfrac{20}{\sqrt{3}}\times 100\right)}(+4+4-4)\times 10^{-6}$$
$$=\dfrac{9\times 10^{9}\times 4\times \sqrt{3}\times 100\times 10^{-6}}{20}$$
$$V_{net}=3.078\times 10^{5}V$$
In the following circuit potential at point 'A' is zero then determine Potential difference across each resistance.
$$\quad As\quad given\quad the\quad potential\quad of\quad A\quad is\quad 0V\quad therefore\quad \\ \quad \quad \quad \quad { V }_{ A }={ V }_{ B }={ V }_{ C }={ V }_{ D }=0V\\ And\quad { V }_{ E }={ V }_{ F }={ V }_{ G }={ V }_{ H }=10V\\ Therefore\quad potential\quad across\quad I\quad resistor\quad becomes\quad \\ \quad \quad 10+5=15V\\ And\quad potential\quad across\quad J\quad resistor\\ \quad 15-10=5V\\ And\quad potential\quad across\quad K\quad resistor\quad becomes\\ \quad 10+5=15V\quad $$
The equivalent capacitance between A and B are:
The equivalent capacitance is $$C$$.
Calculate in the given figure: Charge at $$O$$ is $$10\mu C$$ $$(i)$$ Electric potentials at point $$A$$ and $$B$$, $$(ii)$$ work done in taking the same amount of negative charge from $$A$$ to $$B$$, $$(OA=0.10\ m, OB=0.20\ m$$), assume that the charge at $$O$$ stays fixed
If the difference between the radii of the two sphere of a spherical conductor is increased state whether the capacitance will increase or decrease.
A capacitor of capacitance $$10 \mu F$$ is connected to battery of emf $$20\ V$$. Without disconnecting the source a dielectric $$(K=4)$$ is introduced to fill the space between two plates of the capacitor. Calculate the-
(a) charge before the dielectric was introduced.
(b) charge After the dielectric was introduced.
A nucleus has charged of + $$50$$ e A proton is located at a distance of $$10$$ m The potential at this point in volt will be-
What is meant by saying that the electric potential at a point is $$1\ Volt$$?
What should be the charge on a sphere of radius 2 cm so that when it is brought in contact with another sphere of radius 5 cm carrying a charge of $$10\mu C$$, there is no transfer of charge between the spheres?
A $$ 5.0\mu F$$ capacitor is charged to 12 V. The positive of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires.
A parallel plate air capacitor is connected to a battery.If plates of the capacitor are pulled farther apart, then state whether the following statements are true or flase
a. Strength of electric field inside the capacitor remains unchanged, if battery is disconnected before pulling the plates.
b. During the process, work is done by external force applied to pull the plates irrespective of whether the battery is disconnected or not.
c.Strain energy in the capacitor decreases if the battery remains connected
A parallel-plate capacitor of capacitance $$5 \mu F$$ is connected to a battery of emf $$6 V.$$ The separation between the plates is $$2 mm$$. A dielectric slab of thickness $$1 mm$$ and dielectric constant $$5$$ is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination.
Answer the following questions.
Derive an expression for equivalent capacitance of three capacitors when connected in series.
Derive an expression for equivalent capacitance of three capacitors when connected in parallel.
In a fig (a) Three capacitors of capacitances $$ C_{1} $$ , $$ C_{2} $$ , $$ C_{3} $$ are connected in series between points A and D. In series' first plate of each capacitor has charge +Q and second plate of each capacitor has charge -Q i.e., charge on each capacitor is Q.
Let the potential differences across the capacitors $$ C_{1} $$
$$ C_{2} $$ , $$ C_{3} $$ be $$V_{1} $$ , $$ V_{2} $$ , $$V_{3} $$
respectively. As the second plate of first capacitor $$C_{1} $$ and first plate of second capacitor $$C_{2} $$ are connected together, their potentials are equal.
Let the potential differences across the capacitors $$ C_{1} $$
$$ C_{2} $$ , $$ C_{3} $$ be $$V_{1} $$ , $$ V_{2} $$ , $$V_{3} $$
respectively. As the second plate of first capacitor $$C_{1} $$ and first plate of second capacitor $$C_{2} $$ are connected together, their potentials are equal.
A test charge 'q' is mover without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure.
At which point (of the two) is the electric potential more and why?
Where is the energy stored due to an electric charge?
What is electric potential? Write its formula and unit.
Electric Potential In fig. 3.1 it is shown that in electric field due to a configuration of charges, the work done in carrying a test charge $$ \left(\mathrm{q}_{0}\right) $$ from point $$ \mathrm{A} $$ to $$ \mathrm{B} $$, depends only upon the distance between $$ \mathrm{A} $$ and $$ \mathrm{B} $$ not on the path adopted.
"Electric potential is the factor which decides the direction of flow of electric charge."It is scalar quantity and it is denoted by V. As liquids flow from higher gravitation level to the lower level; heatflows from higher temperature to the lower temperature, similarly charge (positive) flows from higherelectric potential to the lower potential. Negative charge flows from lower electric potential to the higherpotential.
If $$ \mathrm{V}_{\mathrm{A}} $$ and $$ \mathrm{V}_{\mathrm{B}} $$ be the electric potentials at points $$ \mathrm{A} $$ and $$ \mathrm{B} $$ respectively, then the potential difference betweenpoints A and B is given below,$$\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\dfrac{W_{\mathrm{AB}}}{q_0}\ldots(1)$$
Where, $$ W_{A B} $$ is the work done in carrying test charge $$ +q_{0} $$ from $$ A $$ to $$ B $$.
Here we assume that in this process, the test charge does not disturb the source of field, i.e., all the charges remain at their position. If for a displacement, the change in potential energy is $$ \cup_{B}-U_{A} $$, then the potential difference between the points A and B can be given as
$$V_{B}-V_{A}=\dfrac{U_{B}-U_{A}}{q_{0}}=\dfrac{\Delta U}{q_{0}}\ldots (2)$$
Therefore from equations (1) and (2),
$$V_{B}-V_{A}=\dfrac{U_{B}-U_{A}}{q_{0}}=\dfrac{W_{A B}}{q_{0}}\ldots (3)$$
If $$ \mathrm{q}_{0}=+1 \mathrm{C}, $$ then $$ \mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\mathrm{W}_{\mathrm{AB}} $$
"Electric potential is the factor which decides the direction of flow of electric charge."It is scalar quantity and it is denoted by V. As liquids flow from higher gravitation level to the lower level; heatflows from higher temperature to the lower temperature, similarly charge (positive) flows from higherelectric potential to the lower potential. Negative charge flows from lower electric potential to the higherpotential.
If $$ \mathrm{V}_{\mathrm{A}} $$ and $$ \mathrm{V}_{\mathrm{B}} $$ be the electric potentials at points $$ \mathrm{A} $$ and $$ \mathrm{B} $$ respectively, then the potential difference betweenpoints A and B is given below,$$\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\dfrac{W_{\mathrm{AB}}}{q_0}\ldots(1)$$
Where, $$ W_{A B} $$ is the work done in carrying test charge $$ +q_{0} $$ from $$ A $$ to $$ B $$.
Here we assume that in this process, the test charge does not disturb the source of field, i.e., all the charges remain at their position. If for a displacement, the change in potential energy is $$ \cup_{B}-U_{A} $$, then the potential difference between the points A and B can be given as
$$V_{B}-V_{A}=\dfrac{U_{B}-U_{A}}{q_{0}}=\dfrac{\Delta U}{q_{0}}\ldots (2)$$
Therefore from equations (1) and (2),
$$V_{B}-V_{A}=\dfrac{U_{B}-U_{A}}{q_{0}}=\dfrac{W_{A B}}{q_{0}}\ldots (3)$$
If $$ \mathrm{q}_{0}=+1 \mathrm{C}, $$ then $$ \mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\mathrm{W}_{\mathrm{AB}} $$
Conceptual Questions
If you were asked to design a capacitor in which small size and large capacitance were required, what would be the two most important factors in your design?
A parallel-plate capacitor is formed by two discs with a uniform poorly conducting medium between them. The capacitor was initially charged and then disconnected from a voltage source. Neglecting the edge effects, show that there is no magnetic field between capacitor plates.
A charge $$q$$ is distributed uniformly over the volume of a ball of radius $$R$$. Assuming the permittivity to be equal to unity, find:
(a) the electrostatic self-energy of the ball;
(b) the ratio of the energy $$W_{1}$$ stored in the ball to the energy $$W_{2}$$ pervading the surrounding space.
Distinguish between electric potential and electric potential energy.
Two concentric shells of radii $$d$$ and $$2d$$ respectively are placed far away from a parallel plate capacitor. Each plate of the capacitor is a square of side $$d$$ and separation between the plates is $$d/4\pi$$. The charge on the plates are $$q$$ and $$-q$$. Then the keys $$k_1$$ and $$k_2$$ are closed, find the charges on the shells and electric potential energy left in the parallel plate capacitor in equilibrium.
let the new charge on the parallel plate capacitor be $$+q_1$$ and that on the concentric shells be $$(q - q_1)$$, such that the potential differences across the shell would be same as the potential difference across the plates.
$$\phi_{innershell} = \dfrac{(q - q_1)}{4\pi \varepsilon_0(2d)} + \dfrac{-(q - q_1)}{4\pi \varepsilon_0 d} = \dfrac{-(q - q_1)}{4\pi \varepsilon(2d)}$$
$$\phi_{outershell} = \dfrac{(q - q_1)}{4\pi \varepsilon_0(2d)} + \dfrac{-(q - q_1)}{4\pi \varepsilon_0(2d)} = 0$$
Potential difference, $$V_{shells} = \phi{outer} - \phi{inner} = \dfrac{(q - q_1)}{4\pi \varepsilon_0(2d)}$$
for parallel plate, $$V_{plates} = \dfrac{q_1}{4\pi \varepsilon_0 d}$$
$$V_{shells} = V_{plates}$$
$$\dfrac{(q - q_1)}{4\pi \varepsilon_0(2d)} = \dfrac{q_1}{4\pi \varepsilon_0 d}$$
$$\Longrightarrow (q - q_1) = 2q_1$$
$$q_1 = q/3$$
$$\therefore$$ Charge on the outer shell $$= (q - q_1) = \dfrac{2}{3}q$$
Charge on the inner shell $$= 2q/3$$
Energy left in the parallel plate capacitor $$= \dfrac{1}{2}\dfrac{q^2_1}{C} = \dfrac{q^2}{72\in \varepsilon_0 d}$$
$$\phi_{innershell} = \dfrac{(q - q_1)}{4\pi \varepsilon_0(2d)} + \dfrac{-(q - q_1)}{4\pi \varepsilon_0 d} = \dfrac{-(q - q_1)}{4\pi \varepsilon(2d)}$$
$$\phi_{outershell} = \dfrac{(q - q_1)}{4\pi \varepsilon_0(2d)} + \dfrac{-(q - q_1)}{4\pi \varepsilon_0(2d)} = 0$$
Potential difference, $$V_{shells} = \phi{outer} - \phi{inner} = \dfrac{(q - q_1)}{4\pi \varepsilon_0(2d)}$$
for parallel plate, $$V_{plates} = \dfrac{q_1}{4\pi \varepsilon_0 d}$$
$$V_{shells} = V_{plates}$$
$$\dfrac{(q - q_1)}{4\pi \varepsilon_0(2d)} = \dfrac{q_1}{4\pi \varepsilon_0 d}$$
$$\Longrightarrow (q - q_1) = 2q_1$$
$$q_1 = q/3$$
$$\therefore$$ Charge on the outer shell $$= (q - q_1) = \dfrac{2}{3}q$$
Charge on the inner shell $$= 2q/3$$
Energy left in the parallel plate capacitor $$= \dfrac{1}{2}\dfrac{q^2_1}{C} = \dfrac{q^2}{72\in \varepsilon_0 d}$$
A parallel plate capacitor is charged by a battery and the battery remains connected, a dielectric slab is inserted in the space between the plates. Explain what changes if any, occur in the values of the
(i) potential difference between the plates
(ii) electric field between the plates
(ii) energy stored in the capacitor
$$V\rightarrow constant$$ because battery nemans connected.
$$E = \dfrac {E_{0}}{4} \rightarrow k$$ decreases
$$3.U = \dfrac {1}{2} CV^{2}$$
$$U$$ is same but $$C$$ become $$kC$$
So, $$U = \dfrac {1}{2} CV^{2} \cdot k (k$$ increases)
i.e. Energy store $$\uparrow$$.
A glass plate totally fills up the gap between the electrodes of a parallel-plate capacitor whose capacitance in the absence of that glass plate is equal to $$C = 20\ nF$$. The capacitor is connected to a dc voltage source $$V = 100\ V$$. The plate is slowly, and without friction, extracted from the gap. Find the capacitor energy increment and the mechanical work performed in the process of plate extraction.
A system consists of two thin concentric metal shells of radii $$R_{1}$$ and $$R_{2}$$ with corresponding charges $$q_{1}$$ and $$q_{2}$$. Find the self-energy values $$W_{1}$$ and $$W_{2}$$ of each shell, the interaction energy of the shells $$W_{12}$$, and the total electric energy of the system.
A capacitor of capacitance $$C_{1} = 1.0\mu F$$ carrying initially a voltage $$V = 300\ V$$ is connected in parallel with an uncharged capacitor of capacitance $$C_{2} = 2.0\mu F$$. Find the increment of the electric energy of this system by the moment equilibrium is reached. Explain the result obtained.
Electric Potential and Potential Energy Due to Point Charges(23)
A particle with charge $$+q$$ is at the origin. A particle with charge $$-2q$$ is at $$x=2.00 \,m$$ on the x axis.
(a) For what finite value(s) of x is the electric field zero?
(b) For what finite value(s) of x is the electric potential zero?
Find the mean electrostatic potential produced by an electron in the centre of a hydrogen atom if the electron is in the ground state for which the wave function is $$ \psi (r) = A e^{-r/r_1} $$, where $$A$$ is a certain constant, $$r_1$$ is the first Bohr radius.
A point charge $$q = 3.0\mu C$$ is located at the centre of a spherical layer of uniform isotropic dielectric with permittivity $$\epsilon = 3.0$$. The inside radius of the layer is equal to $$a = 250\ mm$$, the outside radius is $$b = 500\ mm$$. Find the electrostatic energy inside the dielectric layer.
The equivalent capacity between $$A$$ and $$B$$ is.
A uniform electric field of $$10 NC^{-1}$$ exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of $$50$$ cm.
A large conducting plane has a surface charge density $$1.0 \times 10^{-4}\ C m^{-2}$$. Find the electrostatic energy stored in a cubical volume of edge $$1.0\ cm$$ in front of the plane.
Find the equivalent capacitance of the system shown in figure between the point $$a$$ and $$b$$.
Above combination is in series and parallal.
$$C_{eq}$$ between a & b
$$C_{eq}= \dfrac{C_1 C_2}{C_1 + C_2} + C_3 + \dfrac{C_1C_2}{C_1 + C_2}$$
$$C_{eq}= C_3 + \dfrac{2C_1 C_2}{C_1 + C_2} $$ ($$\therefore$$ The three are parallel)
An electric field $$\vec{E}=(\vec{i}20+\vec{j}30)NC^{-1}$$ exists in the space. If the potential at the origin is taken to be zero, find the potential at $$(2m, 2m)$$.
A parallel-plate capacitor having plate area $$20\ cm^2$$ and separation between the plates $$1.00\ mm$$ is connected to a battery of $$12.0\ V$$. The plates are pulled apart to increase the separation to $$2.0\ mm$$. How much energy is absorbed by the battery during the process?
Consider the situation shown in the figure. The switch $$S$$ is open for a long time and then closed. Find the work done by the battery.
A parallel-plate capacitor having plate area $$20 cm^2$$ and separation between the plates 1.00 mm is connected to a battery of $$12.0 V$$. The plates are pulled apart to increase the separation to $$2.0 mm$$. Show and justify that no heat is produced during this transfer of charge as the separation is increased.
If the area of each plate is $$A$$ (figure) and the successive separations are $$d,2d$$ and $$3d$$, the equivalent capacitance across $$A$$ and $$B$$ is $$\cfrac { { \varepsilon }_{ 0 }A }{ xd } $$. Find $$x$$ :
The equivalent circuit is shown in figure.
Here, $$\displaystyle C_1=\frac{A\epsilon_0}{d}$$ and $$\displaystyle C_2=\frac{A\epsilon_0}{3d}=\frac{C_1}{3}$$
As they are in series so equivalent capacitance is $$\displaystyle C_{AB}=\frac{C_1C_2}{C_1+C_2}=\frac{C_1(C_1/3)}{C_1+(C_1/3)}=\frac{C_1}{4}=\frac{A\epsilon_0}{4d}$$
Here, $$\displaystyle C_1=\frac{A\epsilon_0}{d}$$ and $$\displaystyle C_2=\frac{A\epsilon_0}{3d}=\frac{C_1}{3}$$
As they are in series so equivalent capacitance is $$\displaystyle C_{AB}=\frac{C_1C_2}{C_1+C_2}=\frac{C_1(C_1/3)}{C_1+(C_1/3)}=\frac{C_1}{4}=\frac{A\epsilon_0}{4d}$$
In an hydrogen atom, the electron revolves around the nucleus in an orbit of radius $$0.53 \times {10^{ - 10}}m.$$ Then the electrical potential produced by the nucleus at the position of the electron is
Two circular plates of radius 0.1 m are used to form a parallel plate capacitor. if displacement current between the plates is $$2\pi$$ ampere, then find magnetic field produced by displacement current 4 cm from the axis of the plates.
Some capacitors each of capacitance $$30\mu F$$ are connected as shown in Fig. Calculate equivalent capacitance between terminals A and B.
$$Q=\dfrac{11}{15}CV$$
Hence, $$Q=C_{eq}V=\dfrac{11}{15}CV$$
$$C_{eq}=\dfrac{11}{15}C=\dfrac{11}{15}30=22\mu C$$
The above figure shows a hemisphere with a charge of $$4.00 \ \mu C$$ distributed uniformly through its volume. The hemisphere lies on an $$xy$$ plane the way half a grapefruit might lie face down on a kitchen table. Point $$P$$ is located on the plane, along a radial line from the hemisphere's center of curvature, at radial distance $$15 \ cm$$. What is the electric potential at point $$P$$ due to the hemisphere?
Two uniformly large parallel thin plates having charge densities $$+\sigma $$ and $$-\sigma $$ are kept in the X-Z plane at a distance 'd' apart. Sketch an equipotential surface due to electric field between the plates. If a particle of mass m and charge '-q' remains stationary between the plates, what is the magnitude and direction of this field?
The equipotential surface is at a distance d/2 from either plate in X-Z plane. For a particle of charge (-q) at rest between the plates, then
Weight mg acts vertically downward
Electric force $$qE$$ vertically upward
so $$mg = qE$$
$$E = \dfrac{mg}{q},$$ Vertically downward, i.e., along (-) Y-axis.
Weight mg acts vertically downward
Electric force $$qE$$ vertically upward
so $$mg = qE$$
$$E = \dfrac{mg}{q},$$ Vertically downward, i.e., along (-) Y-axis.
A $$10 V$$ battery is connected to a series of $$n$$ capacitors, each of capacitance $$2.0 µF$$. If the total stored energy is $$25 µJ$$, what is $$n$$?
Define electric potential.
Define the term electric potential. State its S.I. unit.
Write the S.I. unit of capacitance.
What is an equipotential surface?
Find equivalent capacitance between points $$A$$ and $$B$$ as shown in figure.
At what point on the line joining the two charge is the electric potential zero.
Let the potential at point A be 0.
$$\frac{kq_{1}}{x}+\frac{kq_{2}}{d-x}=0$$
$$\frac{kq_{1}}{x}=-\frac{kq_{2}}{d-x}$$
$$\Rightarrow \frac{q_{1}}{x}=\frac{q_{2}}{x-d}$$
$$\Rightarrow x=\frac{q_{1}d}{q_{1}-q_{2}}$$
Name the physical quantity whose $$SI$$ unit is $$J/C$$. is it a scalar or a vector quantity?
A 12pF capacitor is connected to a 50 V battery . How much of electrostatic energy is stored in the capacitor?
What is the equivalent capacitance of the combination ?
Two condensers A and B each having slabs of dielectric constant $$K = 2$$ are connected in series. When they are connected across 230 V supply, potential across A is 130 V and that across B is 100 V. If the dielectric in the condenser of smaller capacitance is replaced by one for which $$K=5$$ what will be the values of potential difference across them?
What is the charge density on each surface of the dielectric?
Draw a plot showing the variation of (i) electric field $$\left(E\right)$$ and (ii) electric potential $$\left(V\right)$$ with distance $$r$$ due to a point charge $$Q$$.
The electric field due to point charge Q is $$E=kQ/r^2$$ and the potential is $$ V=kQ/r$$
Thus, the graph E,V vs r will be as shown in figure.
If $$100$$ joule of work must be done to move electric charge equal to $$4\ C$$ from a place, where potential is $$-10$$ volt to another place, where potential is V volt, find the value of V.
State whether true or false : The unit of voltage is $$JC^{-1}$$.
A charge of $$20\mu C$$ produces an electric field. Two points are $$10\ cm$$ and $$5\ cm$$ away from this charge. Find the values of potentials at these points and also find the amount of work done to take an electron from one point to the other.
$$V_A=\cfrac{1}{4\pi {\epsilon}_0} \cfrac{q}{r_A} = \cfrac{9 \times {10}^9 \times 20 \times {10}^{-6}}{0.05}$$
$$V_A=36 \times {10}^{-5} \; V$$
$$V_B=\cfrac{1}{4\pi {\epsilon}_0} \cfrac{q}{r_B} = \cfrac{9 \times {10}^9 \times 20 \times {10}^{-6}}{0.1}$$
$$V_B=18 \times {10}^{-5} \; V$$
$$\text{Work done} = e \triangle{V}$$
$$W = 1.6 \times {10}^{-19} (36 \times {10}^{-5} - 18 \times {10}^{-5})$$
$$W=28.8 \times {10}^{-24} j = 0.18meV$$
$$ABCD$$ is a square of side $$0.2\ m$$. Charges of $$2\times 10^{-9}, 4\times 10^{-9}, 8\times 10^{-9}$$ coulomb are placed at the corners A, B and C respectively. Calculate the work required to transfer a charge of $$2\times 10^{-9}$$ coulomb from corner D to the centre of the square.
$$AC=BD=\sqrt{(0.2)^{2}+(0.2)^{2}}$$
$$=\sqrt{2}\times 0.2=0.28\; m$$
$$\therefore \: AO=BO=CO=\dfrac{0.28\;m}{2}=0.14\; m$$
$$V$$ i.e, Potential at $$O; \: V_{o}=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{q_{1}}{AO}+\dfrac{q_{2}}{BO}+\dfrac{q_{3}}{CO} \right ]$$
$$V$$ i.e, Potential at $$D; \: V_{D}=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{q_{1}}{AD}+\dfrac{q_{2}}{BD}+\dfrac{q_{3}}{CD} \right ]$$
Work done $$=q[V_{o}-V_{D}]$$
$$=\dfrac{2\times 10^{-9}}{4\pi \varepsilon _{0}}\left [ \dfrac{2\times 10^{-9}}{0.14}+\dfrac{4\times 10^{-9}}{0.14}+\dfrac{8\times 10^{-9}}{0.14}- \dfrac{2\times 10^{-9}}{0.2}-\dfrac{4\times 10^{-9}}{0.28}-\dfrac{8\times 10^{-9}}{0.2}\right ]$$
On solving, we get
$$W=2\times 10^{-9}\times 2\times 10^{9}\left [ \dfrac{14\times 10^{-9}}{0.14}-\dfrac{10\times 10^{-9}}{0.2}-\dfrac{4\times 10^{-9}}{0.28} \right ]$$
$$=4\times \left [ 10^{-7}-0.5\times 10^{-7}-0.14\times 10^{-7} \right ]$$
$$=5.40\times 10^{-7}$$ Joules.
What is the nature of the charge on the plates of capacitor ?
The earth has Volume 'V' and surface area 'A' then its capacitance would be
Calculate the electrostatic potential energy for a system of three point charges placed at the corners of an equilateral triangle of side 'a'.
Find expression for electric potential due to point Q at distance r
In the above figure (a), we move an electron from an infinite distance to a point at distance $$R=8.00 \ cm$$ from a tiny charged ball. The move requires work $$W=2.16 \times 10^{-13} \ J$$ by us.
(a) What is the charge $$Q$$ on the ball? In figure (b), the ball has been sliced up and the slices spread out so that an equal amount of charge is at the hour positions on a circular clock face of radius $$R=8.00 \ cm$$. Now the electron is brought from an infinite distance to the center of the circle.
(b) With that addition of the electron to the system of $$12$$ charged particles, what is the change in the electric potential energy of the system?
A spherical drop of capacitance $$12\mu F$$ is broken into eight drops of equal radius. Then, what is the capacitance of each small drop in $$\mu F$$?
Draw $$3$$ equipotential surface corresponding to a field that uniformly increases in magnitude but remains constant along Z-direction. How are these surface different from that of a constant electric field along Z-direction?
For constant electric field $$\underset{E}{\rightarrow}$$ For increasing electric field
Difference : For constant electric field, the equipotential surface are equidistant for same potential difference between surfaces; while for increasing electric field, the separation between these surfaces decreases, in the direction of increasing field, for the same potential difference between them.
Difference : For constant electric field, the equipotential surface are equidistant for same potential difference between surfaces; while for increasing electric field, the separation between these surfaces decreases, in the direction of increasing field, for the same potential difference between them.
Three identical capacitors $$C_1, C_2$$ and $$C_3$$ have a capacitance of $$1.0 \mu F$$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $$C_1$$ is then filled completely with a dielectric material of relative permittivity $$\in_r$$. The cell electromotive force (emf) $$V_o = 8V$$. First, the switch $$S_1$$ is closed while the switch $$S_2$$ is kept open. When the capacitor $$C_3$$ is fully charged, $$S_1$$ is opened and $$S_2$$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $$C_3$$ is found to be $$5\mu C$$. The value of $$4\times \in_r$$ is:
The capacitor $$C_3$$ charges to $$8 \mu C$$ as the switch $$S_1$$ is closed.
Ref. image 1
But when switch $$S_1$$ is opened and $$S_2$$ is closed, then capacitor $$C_3$$ is charged to $$5 \mu C$$, thus the net charge $$8 \mu C - 5 \mu C = 3 \mu C$$ resides on $$C_1$$ and $$C_2$$, as they are connected in series.
Ref. image 2
So, Apply the Kirchhoff loop,
$$\dfrac{q_1}{C_1} + \dfrac{q_2}{C_1 \varepsilon_r} = \dfrac{q_3}{C_3}$$
$$\dfrac{q}{C} \left(1 + \dfrac{1}{\varepsilon_r} \right) = \dfrac{5}{1}$$
$$3 \left(1 + \dfrac{1}{\varepsilon_r}\right) = 5$$
$$\varepsilon_r = 1.5$$
Show that the effective capacitance, C, of a series combination, of three capacitors, $$C_1, C_2$$ and $$C_3$$ is given by
$$C=\frac {C_1C_2C_3}{(C_1C_2+C_2C_3+C_3C_1)}$$
Three capacitors each of capacitance 9 pF are connected in series as shown in figure.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor, if the combination is connected to a 120 volt supply?
A parallel plate capacitor has plates with area $$A$$ & separation $$d$$. A battery charges the plates to a potential difference of $$V_0$$. The battery is then disconnected & a di-electric slab of constant $$K$$ & thickness $$d$$ is introduced. Calculate the positive work done by the system (capacitor + slab) on the man who introduces the slab.
Identical charges q each are placed at the eight corners of a cube of side b. Find the electrostatic potential energy of a charge +q placed at the center of the cube
A charge of 5$$\mu$$ C is placed at the center of a square ABCD of side 10 cm. Find the work done (in $$\mu$$J) in moving a charge of 1$$\mu$$C from A to B.
For a square ABCD, $$OA=OB=OC=OD=r$$
As charge $$5 \mu C$$ is placed on center O so every point A,B,C and D have same potential $$V=5k/r$$.
Thus potential difference $$(dV)$$ between A and B will be zero.
Work done $$W=qdV=1(0)=0$$
As charge $$5 \mu C$$ is placed on center O so every point A,B,C and D have same potential $$V=5k/r$$.
Thus potential difference $$(dV)$$ between A and B will be zero.
Work done $$W=qdV=1(0)=0$$
At a point due to a point charge, the values of electric field intensity and potential are 32 N$$C^{-1}$$ and 16 J $$C^{-1}$$ ,respectively. Calculate the
a. magnitude of the charge,and
b. distance of the charge from the point of observation.
How much energy in kJ will be stored by a capacitor of 470 $$\mu F$$ when charged by a battery of 20 V?
A parallel plate capacitor is charged by a battery.The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates. How will (i) its capacitance, (ii) electric field between the plates and (iii) energy stored in the capacitor be affected ? Justify your answer giving necessary mathematical expressions for each case.
A parallel-plate capacitor has the air as a medium between its plates and its capacitance is C. A slab of the dielectric constant K is introduced so as to fill the left half of space of the capacitor, as shown. What is the new capacitance of the capacitor?
What is the work done in moving a charge of $$50 nC$$ between two points on an equipotential surface?
An uncharged capacitor C is connected to a battery through a resistance R. Show that by the time the capacitor gets fully charged, the energy dissipated in R is the same as the energy stored in C.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =$$ 10^{-12}$$ F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Two charges $$2\mu C$$ and $$-2\mu C$$ are placed at points A and B 6 cm apart.(a) Identify an equipotential surface of the system.(b) What is the direction of the electric field at every point on this surface?
Three capacitors each of capacitance $$9 pF$$ are connected in series.(a) What is the total capacitance of the combination?(b) What is the potential difference across each capacitor if the combination is connected to a $$120 V$$ supply?
Exercise: In a parallel plate capacitor with air between the plates, each plate has an area of $$6\times 10^{-3} m^{2}$$ and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Explain what would happen if in the capacitor given in Exercise a $$3 mm$$ thick mica sheet (of dielectric constant = 6) were inserted between the plates,(a) While the voltage supply remained connected.(b) After the supply was disconnected.
(a) While the voltage supply remained connected.
(b) After the supply was disconnected.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of $$3.5 \mu C$$. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
An infinite number of charges, each equal to q, are placed along the x-axis at $$x=1,\,x=2,\,x=4,\,x=8$$ and so on. What is the potential at $$x=0$$ due to this set of charges?
Write a relation for polarisation $$\overrightarrow{P}$$ of a dielectric material in the presence of an external electric field $$\overrightarrow{E}$$.
joule coulomb$$^{-1}$$ is same as ___________.
Draw a plot showing the variation of (i) electric field $$\left(E\right)$$ and (ii) electric potential $$\left(V\right)$$ with distance $$r$$ due to a point charge $$Q$$.
The field at distance r due to point charge Q is $$E=kQ/r^2$$ and the potential is $$V=kQ/r$$
The variation of field E and potential V with distance is as shown in figure.
Obtain an expression for equivalent capacitance when three capacitors $$C_1, C_2$$ and $$C_3$$ are connected in series.
Consider that three capacitors having capacitance $$C_1, C_2$$ and $$C_3$$ are connected in series to each other shown in the figure above,
Consider that $$+q$$ charge is given to first plat of first capacitor. By induction a charge $$(-q)$$ will produce at internal surface of second plate of this capacitor and so on.
Consider that potential between the plates of capacitor are $$C_1, C_2$$ and $$C_3$$ respectively so,
$$V_1=\dfrac{q}{C_1}, \ V_2=\dfrac{q}{C_2}$$ and $$V_3=\dfrac{q}{C_3}$$
Now $$V=V_1+V_2+V_3$$
$$\therefore$$ $$V=\dfrac{q}{C_1}+\dfrac{q}{C_2}+\dfrac{q}{C_3}$$ ...(i)
If equivalent capacitance of combination of capacitors is $$C$$ then
$$V=\dfrac{q}{C}$$ ...(ii)
By equations (i) and (ii), we get
$$\dfrac{q}{C}=\dfrac{q}{C_1}+\dfrac{q}{C_2}+\dfrac{q}{C_3}$$
$$\implies $$ $$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$$
Consider that $$+q$$ charge is given to first plat of first capacitor. By induction a charge $$(-q)$$ will produce at internal surface of second plate of this capacitor and so on.
Consider that potential between the plates of capacitor are $$C_1, C_2$$ and $$C_3$$ respectively so,
$$V_1=\dfrac{q}{C_1}, \ V_2=\dfrac{q}{C_2}$$ and $$V_3=\dfrac{q}{C_3}$$
Now $$V=V_1+V_2+V_3$$
$$\therefore$$ $$V=\dfrac{q}{C_1}+\dfrac{q}{C_2}+\dfrac{q}{C_3}$$ ...(i)
If equivalent capacitance of combination of capacitors is $$C$$ then
$$V=\dfrac{q}{C}$$ ...(ii)
By equations (i) and (ii), we get
$$\dfrac{q}{C}=\dfrac{q}{C_1}+\dfrac{q}{C_2}+\dfrac{q}{C_3}$$
$$\implies $$ $$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$$
Obtain an expression for equivalent capacitance C, when three capacitors having capacitance $$C_1, C_2$$ and $$C_3$$ are connected m series.
Let three capacitors $$C_1, C_2$$ and $$C_3$$ are connected in series. We know charge remains same in series combination.
$$V_1=\displaystyle\frac{q}{C_1}$$
$$V_2=\displaystyle\frac{q}{C_2}$$
$$V_3=\displaystyle\frac{q}{C_3}$$
$$V=V_1+V_2+V_3=q\left(\displaystyle\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)$$
If the $$3$$ capacitors are replaced by a single capacitor such that on giving it charge q is capacitance is C then $$V=\displaystyle\frac{q}{C}$$
$$\therefore \displaystyle\frac{q}{C}=q\left(\displaystyle\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)$$
or $$\displaystyle\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$
$$V_1=\displaystyle\frac{q}{C_1}$$
$$V_2=\displaystyle\frac{q}{C_2}$$
$$V_3=\displaystyle\frac{q}{C_3}$$
$$V=V_1+V_2+V_3=q\left(\displaystyle\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)$$
If the $$3$$ capacitors are replaced by a single capacitor such that on giving it charge q is capacitance is C then $$V=\displaystyle\frac{q}{C}$$
$$\therefore \displaystyle\frac{q}{C}=q\left(\displaystyle\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)$$
or $$\displaystyle\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$
Calculate the potential at a point due to a charge of $$3\times {10}^{-9}C$$ located $$9\times {10}^{-2}m$$ away from it.
Calculate the potential at a point due to a charge of $$4\times 10^{-7}C$$ located at $$0.09\ m$$ away.
Two point charges $$5 \times 10^{-8} C$$ and $$-3 \times 10^{-8} C$$ are located 16 cm apart. At what point on the line joining these charges the electric potential will be zero?
Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.
Given diagram show parallel plate capacitor completely filled with dielectric.
Define electric capacity. Derive an expression for the capacitance of a parallel plate capacitor in which a dielectric medium of dielectric constant $$K$$ fills the space between the plates. Draw the necessary diagram.
Capacitance or electric capacity indicates the ability of a system to store charge. It is defined as the ratio of charge stored in the conductors to the potential difference across the conductors. Its SI unit is Farad ($$F$$). Hence:
$$C = \cfrac{Q}{V}$$
To derive the expression of capacitance of parallel plate capacitor of conductor area $$A$$ and distance between the conductors as $$d$$ with a dielectric of dielectric constant $$K$$ carrying a charge $$Q$$ and potential difference $$V$$, refer to the attached figure.
Electric field between plates of the capacitor can be found using Gauss Law.
$$E = \cfrac{\sigma}{K\varepsilon_o }$$
$$E = \cfrac{Q}{K\varepsilon_o A }$$ as $$\sigma = \cfrac{Q}{A}$$
Electric field is defined as the gradient of potential.
$$E = \cfrac{V}{d}$$
$$\implies V = \cfrac{Qd}{K\varepsilon_o A }$$
By definition of capacitance,
$$C = \cfrac{Q}{V}$$
$$=\cfrac{K\varepsilon_o A }{d}$$
Write the definition of electric potential. Calculate the electric potential due to a point charge Q at a distance r from it. Draw a graph between electric potential V and distance r for a point charge Q.
Electric potential is the amount of electric potential energy that a unitary point electric charge would have if located at any point in space, and is equal to the work done by an external agent in carrying a unit of positive charge from the arbitrarily chosen reference point (usually infinity) to that point without any acceleration.
Electric potential due to point charge Q at a distance r is
$$V=-\int_{\infty}^{r}Edr$$
$$=\int_{-\infty}^{r}\dfrac{Q}{4\pi \epsilon_0 r^2}dr$$
$$=\dfrac{Q}{4\pi \epsilon_0 r}$$
Deduce an expression for the effective capacitance of capacitors of $$C_{1}, C_{2}$$ and $$C_{3}$$ connected in series
Capacitors in series: Consider three capacitors of capacitance $$C_{1}, C_{2}$$ and $$C_{3}$$ connected in series. Let $$V$$ be the potential difference applied across the series combination. Each capacitor carries the same amount of charge $$q$$. Let $$V_{1}, V_{2}, V_{3}$$ be the potential difference across the capacitors $$C_{1}, C_{2}, C_{3}$$ respectively. Thus $$V = V_{1} + V_{2} + V_{3}$$.
The potential difference across each capacitor is,
$$V_{1} = \dfrac {q}{C_{1}}; V_{2} = \dfrac {q}{C_{2}}; V_{3} = \dfrac {q}{C_{3}}$$;
$$V = \dfrac {q}{C_{1}} + \dfrac {q}{C_{2}} +\dfrac {q}{C_{3}} = q\left [\dfrac {1}{C_{1}} +\dfrac {1}{C_{2}} + \dfrac {1}{C_{3}}\right ]$$
If $$C_S$$ be the effective capacitance of the series combination, it should acquire a charge $$q$$ when a voltage $$V$$ is applied across it.
$$V = \dfrac {q}{C_{S}} = \dfrac {q}{C_{1}} + \dfrac {q}{C_{2}} + \dfrac {q}{C_{3}}$$
$$\therefore \dfrac {1}{C_{S}} = \dfrac {1}{C_{1}} + \dfrac {1}{C_{2}} + \dfrac {1}{C_{3}}$$
The potential difference across each capacitor is,
$$V_{1} = \dfrac {q}{C_{1}}; V_{2} = \dfrac {q}{C_{2}}; V_{3} = \dfrac {q}{C_{3}}$$;
$$V = \dfrac {q}{C_{1}} + \dfrac {q}{C_{2}} +\dfrac {q}{C_{3}} = q\left [\dfrac {1}{C_{1}} +\dfrac {1}{C_{2}} + \dfrac {1}{C_{3}}\right ]$$
If $$C_S$$ be the effective capacitance of the series combination, it should acquire a charge $$q$$ when a voltage $$V$$ is applied across it.
$$V = \dfrac {q}{C_{S}} = \dfrac {q}{C_{1}} + \dfrac {q}{C_{2}} + \dfrac {q}{C_{3}}$$
$$\therefore \dfrac {1}{C_{S}} = \dfrac {1}{C_{1}} + \dfrac {1}{C_{2}} + \dfrac {1}{C_{3}}$$
Derive the formula for equivalent capacitance when the capacitors are connected in series.
Let the capacitance of each capacitor be $$C_1,$$ $$C_2$$ and $$C_3$$ and their equivalent capacitance be $$C_{eq}$$.
As these capacitors are connected in series, thus charge across each capacitor is same as $$Q$$.
When some electrical components, let say 3, are connected in series with each other, the potential difference of the battery $$V$$ gets divided across each component as $$V_1,$$ $$V_2$$ and $$V_3$$ as shown in the figure.$$\therefore$$ $$V = V_1 +V_2 + V_3$$
Using $$V = \dfrac{Q}{C}$$
We get $$\dfrac{Q}{C_{eq}} = \dfrac{Q}{C_1}+\dfrac{Q}{C_2}+\dfrac{Q}{C_3}$$
$$\implies$$ Equivalent capacitance for series combination $$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$$
In general, $$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+ ..........+\dfrac{1}{C_n}$$
Derive the expression for the capacitance of a parallel plate capacitor.
$$\textbf{Step 1 - Draw figure and find electric field}$$
Let the two plates be parallel to each other, each carrying
a surface charge density $$+\sigma$$ and $$-\sigma$$ respectively.
$$A$$ is the area of plates and $$d$$ is the separation between them
The electric field of a thin charged plate is given by
$$E = \dfrac {\sigma}{2\varepsilon_{0}}$$ and is directed normally outward from the plate.
The total electric field between the two places is given as
$$\vec {E} = \vec {E}_{A} + \vec {E}_{B}$$
$$\vec {E} = \dfrac {\sigma}{2\varepsilon_{0}} \hat {i} + \dfrac {-\sigma}{2\varepsilon_{0}} (-\hat {i}) = \dfrac {\sigma}{\varepsilon_{0}} \hat {i}$$
$$E = |\vec {E}| = \dfrac {\sigma}{\varepsilon_{0}} = \dfrac {\dfrac {Q}{A}}{\varepsilon_{0}} = \dfrac {Q}{A\varepsilon_{0}}$$
$$\textbf{Step 2 - Potential difference between the two plates}$$
Potential difference $$V = E\times d = \dfrac {Qd}{A \varepsilon_{0}}$$
$$\textbf{Step 3 - Capacitance}$$
The capacitance is
$$C = \dfrac {Q}{V} = \dfrac {Q}{\dfrac {Qd}{A\varepsilon_{0}}} = \dfrac {\varepsilon_{0}A}{d}$$
In a parallel plate capacitor with increase in distance between the plates, its capacity ................
Derive an expression for the electric potential due to a point charge :
TWO point charges $$+2nc$$ and $$-4nc$$ are $$1\ m$$ apart in air. Find the positions along the line joining the two charges at which resultant potential is zero.
Let the potential at point P is zero which is situated at a distance $$x$$ to the left of $$1nC$$ charge as shown in the figure.
Potential due a charge $$q$$ is given by $$V = \dfrac{kq}{r}$$ where $$k = \dfrac{1}{4\pi \epsilon_o}$$
Thus potential at point P $$V_p = \dfrac{k(1nC)}{x} - \dfrac{k (4nC)}{1+x}$$
But $$V_p =0$$
$$\therefore$$ $$ \dfrac{k(1nC)}{x} - \dfrac{k (4nC)}{1+x}=0$$
Or $$ \dfrac{1}{x} - \dfrac{4}{1+x}=0$$
Or $$ \dfrac{1}{x} = \dfrac{4}{1+x}$$
Or $$3x = 1$$
$$\implies$$ $$x =0.33 m$$
Two identified parallel plate capacitors $$A$$ and $$B$$ are connected to a battery of $$V$$ volts with the switch $$S$$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $$K$$. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
A network of four $$10 \mu F$$ capacitors is connected to a $$500V$$ supply, as shown in fig. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor.(Note, the charge on a capacitor is the charge on the place with higher potential, equal and opposite to the charge on the plate with lower potential).
$$C_1 = C_2 = C_3 = C_4 = 10 \mu F$$
$$C_1 , C_2$$ and $$C_3$$ are in series
So
$$\dfrac{1}{C'} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}$$
$$\Rightarrow C' = \dfrac{10}{3} \mu F$$
$$C_{AD} = C_{eq} = C' + C_4$$ [$$\because$$ They are in parallel]
$$\Rightarrow C_{eq} = \dfrac{10}{3} + 10 = \dfrac{40}{3} \mu F$$
Charge on $$C_4 = 10 \mu F \times 500 v = 5 \times 10^{-3} C$$
Charge of $$C_1 , C_2$$ and $$C_3$$ will be same as they are in series
So charge on $$C_1 , C_2$$ and $$C_3$$
$$= C' \times V = \dfrac{10}{3} \mu F \times 500 = \dfrac{5}{3} \times 10^{-3} C$$
Each of the plates shown in the figure has surface area $$( 96/ \epsilon_0) \times 10^{-12}\ m^2$$ on one side and the separation between the consecutive plates is $$4.0\ mm$$. The emf of the battery connected is $$10\ \text{volts}$$. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.
Here three capacitors are formed
And each of
Surface area of the plates is $$A = \dfrac{96}{\varepsilon_0} \times 10^{-12} f.m$$
The distance between the plates is $$d = 4 mm = 4 \times 10^{-3} m$$
$$\therefore$$ Capacitance of a capacitor
$$C = \dfrac{\varepsilon_0 A}{d} = \dfrac{\varepsilon_0 \dfrac{96 \times 10^{-12}}{\varepsilon_0}}{4 \times 10^{-3}} = 24 \times 10^{-9} F$$
$$\therefore$$ As three capacitor are arranged is series
So, $$C_{eq} = \dfrac{C}{q} = \dfrac{24 \times 10^{-9}}{3} = 8 \times 10^{-9}$$
$$\therefore $$ The total charge to a capacitor
$$q=CV$$
$$= 8 \times 10^{-9} \times 10 =8 \times 10^{-8} C$$
$$\therefore$$ The charge of a single Plate $$ q'= 2 \times 8 \times 10^{-8} = 16 \times 10^{-8} = 0.16 \times 10^{-6} = 0.16 \mu C$$
Predict the polarity of the capacitor in the situation described by figure
It is required to construct a $$10 \mu F$$ capacitor which can be connected across a $$200V$$ battery. Capacitors of capacitance $$10\mu F$$ are available but they can withstand only $$50V$$ Design a combination which can yield the desired result
Identify and explain about the diode whose characteristics curve is:
The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $$C_{0}$$ and $$W_{0}$$. If the air between the plates is replaced by glass (dielectric constant$$=5$$) find the capacitance of the consider and the energy stored in it.
Three identical capacitors $$C1, C2$$ and $$C3$$ of capacitance $$6\ mF$$ each are connected to a $$12\ V$$ battery as shown
Find:
charge on each capacitor
equivalent capacitance of the network
energy stored in the network of capacitors
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness t (t<d). Find the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric.
Where Given
k = Dielectric constant
t = thickness of dielectric
Here; the net electric field experienced by the dielectric is
$$\vec{E} = \vec{E}_0 - \vec{E}_p $$ where $$\vec{E_0}$$. ___(1)
the applied E. Field and $$\vec{E_p}$$ is the electric field that is produced inside the dielectric due to polarization of charges.
Potential ditlerence between the plates :
$$V = E_0 (d - t) + Et.$$ __(2) $$\because E = \dfrac{dv}{dx}$$
But $$E = \dfrac{E_0}{k}$$ __(3)
from (2) and (3)
$$\Rightarrow V = E_0 \left[d - t + \dfrac{t}{k} \right]$$ __(4)
$$E_0 = \dfrac{\varepsilon}{\varepsilon_0} = \dfrac{Q}{A \varepsilon_0}$$ __(5)
$$V = \dfrac{Q}{A \varepsilon_0} \left[d - t + \dfrac{t}{k} \right]$$ __(6)
Since $$Q = CV \Rightarrow C = \dfrac{V}{Q} $$ __(7)
From (6) and (7)
$$C_k = \dfrac{\varepsilon_0 A}{\left(d - t + \dfrac{t}{k} \right)} $$ and $$C = \dfrac{\varepsilon_0 A}{d}$$
Hence,
the ratio of the two:-
$$\dfrac{Ck}{C} = \dfrac{\dfrac{\varepsilon_0 A}{\left(d - t + \dfrac{t}{k}\right)}}{\dfrac{\varepsilon_0 A}{d}}$$
$$= \dfrac{d}{\left(d - t + \dfrac{t}{k} \right)}$$
A proton is moved 15 cm on path parallel to the field lines of a uniform electric field of $$2.0\times 10^{5}$$ V/m
What are the possible change in potential ?consider both cases of a moving the proton .
In a conductor, a point P is at a higher potential than another point Q. In which direction do the electrons move?
Find the potential energy of an electric dipole (length 4 cm) each experiences a torque of $$ 4\sqrt { 3 } $$ Nm placed in a from electric field angle $$ 60^\circ q = 1 + nc. $$
Derive the expression for the capacitance of a parallel plate capacitore having plate area A and plate separation d.
Let a parallel plate capacitor be applied a voltage $$V$$ across its terminals due to which a charge $$Q$$ develops across its ends.
Separation between plates is $$d$$ and area of plates is $$A$$.
From gauss's theorem one can show electric field inside the capaciton plates $$Q$$,
$$E = \dfrac{\sigma}{\epsilon_o}$$, where $$\sigma = \dfrac{Q}{A}$$ (charge / Area)
$$\Rightarrow E = \dfrac{Q}{A \epsilon_o}$$
Now, voltage across the plates is related by :
$$V = Ed$$ [In general , $$dV = - \vec{E}. d \vec{r}$$ but we take as $$\vec{E}$$ is uniform]
$$\Rightarrow V = \dfrac{Qd}{A \epsilon_o} \Rightarrow \dfrac{Q}{V} = \dfrac{A \epsilon_o}{d}$$
Now, capacitance is defined by $$C = \dfrac{Q}{V}$$, thus we get :-
$$C = \dfrac{A \epsilon_o}{d} $$
An infinite number of electric charges each equal to 2 nano coulombs in magnitude are placed along x-axis at x=1 cm, x= 3cm , x= 9 cm , x= 27 cm ...... and so on. In these setup if the consecutive charges have opposite sign, then the electric potential at x = 0 will be?
$$\textbf{Step 1 - Potential due to a charge, Refer Figure}$$
$$V = \dfrac {kq}{r}$$
$$\textbf{Step 2 - At x = 0, potential due to all charges}$$
$$V = \dfrac {kq}{0.01} - \dfrac {kq}{0.03} + \dfrac {kq}{0.09} - \dfrac {kq}{0.27} + .....$$
$$= kq \left [100 - \dfrac {100}{3} + \dfrac {100}{9} + \dfrac {100}{27} + .....\right ]$$
$$= 100 \times 9\times 10^{9} \times 2\times 10^{-9} \left [1 - \dfrac {1}{3} + \dfrac {1}{9} - \dfrac {1}{27} + \dfrac {1}{81} .....\right ]$$
$$= 100 \times 9\times 10^{-9 + 9} \times 2 \left [ \dfrac {1}{1 - \dfrac {1}{3}}\right ]$$
$$= 100 \times 18 \left [\dfrac {3}{3 - 1}\right ]$$
$$= 100 \times 18 \left [\dfrac {3}{2}\right ]$$
$$V = 2700\ volt$$