Electrostatic Potential And Capacitance - Class 12 Engineering Physics - Extra Questions

What is the field in the cavity if a conductor having a cavity is charged? Does the result depend on the shape and size of cavity or conductor?

Zero field exists inside the cavity of a conductor. As the charge resides on the outer surface of a conductor irrespective of the shape/size of the conductor and the field inside the cavity of the is always zero .


Define an equipotential surface. Draw equipotential surfaces: Can electric field exist tangential to an equipotential surface? Give reason.

Equipotential surface : An equipotential surface is the surface with a constant value of potential at all points on the surface. 
No, if the field lines are tangential, work will be done in moving a charge on the surface which goes agains the definition of equipotential surface.
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When a small uncharged conducting ball of radius a =1cm and mass m = 50 g is dropped from a height h above the center of another large conducting sphere of radius b (=1 m) having charge Q (=100 $$\mu$$C), it rises to a height $$h_1$$ (= 2m) after the collision. The value of h is $$\dfrac{x}{10}$$. Find x. 
Assume that during the impact there is no dissipation of energy.

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Potential of bigger sphere is $$\displaystyle V = \frac{kQ}{b}$$

When small ball comes in contact with bigger, their potential will becomes same. So

$$\displaystyle \frac{kq}{a} = \frac{k(Q-q)}{b}$$

Neglecting q in comparison to Q, we get

$$\displaystyle  q = \frac{Qa}{b} = 1 \mu C$$

Apply conservation of energy after collision 

$$\displaystyle \frac{1}{2} m \left(\sqrt{2gh}\right)^2 + \frac{kqQ}{b} = mgh_1 + \frac{kqQ}{b+h_1}$$

or $$\displaystyle h = \frac{4}{5}m$$



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 $$2.0 mm$$. Calculate the stored energy in the electric field before and after the process.

area $$=a = 20 cm^2 = 2 \times 10^{-2} m^2$$
$$d = $$ separation $$= 1mm = 10^{-3} m$$
$$Ci = \dfrac{\varepsilon_0 \times 2 \times 10^{-3}}{10^{-3}} = 2 \varepsilon_0$$     $$Cf = \dfrac{\varepsilon_0 \times 2 \times 10^{-3}}{2 \times 10^{-3}} = \varepsilon_0$$

Before the process
$$E_i = (1/2) \times Ci \times v^2 = (1/2) \times 2 \times 8.85 \times 10^{-12} \times 144 = 12.7 \times 10^{-10} J$$

After the force
$$E_i = (1/2) \times Cf \times v^2 = (1/2) \times 8.85 \times 10^{-12} \times 144 = 6.35 \times 10^{-10}J$$


Why are equipotential surfaces perpendicular to field lines ?

Consider if they are not perpendicular, there will be two components which can cause charge to move along the surface. This means that there is some potential difference which is cannot be true since equipotential surfaces, by definition, should be a surface where the potential is constant accross the surface. 


Write two properties of equipotential surfaces. Depict equipotential surfaces due to an isolated point charge. Why do the equipotential surfaces get closer as the distance between the equipotential surface and the source charge decreases ?

Properties of equipotential surfaces:
1. The electric potential value is same across the surface. 
2. The electric field is directed normal to the equipotential surface at all points.

For an isolated point charge electric potential is given by
$$V=\frac{kQ}{r}$$ 
So spheres centered on the charge forms an equipotential surface.

Now, $$E=-\frac{dV}{dr}$$ 
$$dr = \frac{-dV}{E}$$
But dV is constant for an equipotential surface. Therefore, 
$$dr\propto\frac{1}{E}$$
Hence, the equipotential surfaces get closer as the distance between the equipotential surface and the source charge decreases.


A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?


$$\textbf{Step 1 : Formula of Energy stored in capacitor}$$
The Energy stored $$U$$ in a capacitor of Capacitance $$C$$, Charge $$Q$$ and Potential Difference $$V$$ across it is given by:

            $$U = \cfrac 12 CV^2 = \cfrac {Q^2}{2C} = \cfrac 12 QV$$               $$....(1)$$

$$\textbf{Step 2: Calculation of Energy stored}$$
Given, Capacitance $$C = 12pF = 12 \times 10^{-12} F$$
And Potential difference $$V= 50V$$

So using the first formula from Equation $$(1)$$, we get Energy stored as:
                 $$U=\dfrac{1}{2}CV^2=\dfrac{1}{2}\times [12\times 10^{-12}F]\times [(50 V)^2]=15\times 10^{-9}\,J$$

                 $$\boxed{U=15\,nJ}$$

Hence Energy stored in capacitor is $$15\,nJ$$.


Two capacitors of unknown capacitances $$\displaystyle { C }_{ 1 }$$ and $$\displaystyle { C }_{ 2 }$$ are connected first in series and then in parallel across a battery of 100V. If the energy stored in two combinations is 0.045 J and 0.25 J respectively, determine the value of $$\displaystyle { C }_{ 1 }$$ and $$\displaystyle { C }_{ 2 }$$. Also calculate the charge on each capacitor in parallel combination.

When the capacitor are connected in parallel,
Equivalent Capacitance,
$$C_P=C_1+C_2$$
The energy stored in the combination
$$E_P=\cfrac{1}{2}C_PV^2$$
$$\Rightarrow 0.25=\cfrac{1}{2}\left(C_1+C_2\right)(100)^2$$
$$\Rightarrow C_1+C_2=5\times 10^{-5}\longrightarrow(i)$$
Similarly when the capacitor are connected in series
$$C_s=\cfrac{C_1C_2}{C_1+C_2}$$
Energy Stored $$E_s=\cfrac{1}{2}C_sV^2=\cfrac{1}{2}\cfrac{C_1C_2}{C_1+C_2}(100)^2=0.045$$
$$\Rightarrow C_1C_2=4.5\times 10^{-10}$$

Now, $$\left(C_1-C_2\right)^2=\left(C_1+C_2\right)^2-4C_1C_2$$
$$=\left(5\times 10^{-5}\right)^2-4\left(4.5\times 10^{-10}\right)$$
$$=7 \times 10^{-10}$$
$$C_1-C_2=\sqrt{7\times 10^{-10}}=2.64 \times 10^{-5}\longrightarrow(ii)$$
on Solving $$(i)$$ and $$(ii)$$ we get,
$$C_1=38 \mu F$$,                 $$C_2=1.2\mu F$$
charges 
$$Q_1=C_1V=38\times 10^{-6}\times 100=38\times 10^{-4}C$$
$$Q_2=C_2V=12\times 10^{-6}\times 100=12\times 10^{-4}C$$


For any charge configuration, equipotential surface through a point is normal to the electric field. Justify :

For any charge configuration, equipotential surface through a point is normal to the electric field. We will prove this statement using contradiction.
So, if we assume the statement to be false, there can be a component of electric field along the equipotential surface. Now, a non-zero work is done on a test charge moved along the equipotential surface given by $$dW = qEdx$$. But work done on the test charge on movement through an equipotential surface is zero. Hence, the assumption was wrong. Thus, the electric field cannot have any component along the equipotential surface which implies that it has to be perpendicular to the equipotential surface.


A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Hint:- Use the charge conservation to find final charge and final potential and hence calculate ration of energies .

$$\textbf{Step 1: Initial energy stored in capacitor }$$
                $$U_i = \dfrac{1}{2} CV^2$$                                       $$....(1)$$

$$\textbf{Step 2: Calculate charges on both capacitor after they are connected}$$

Let the Initial charge be $$Q$$
After connecting with another capacitor of the equal capacitance, the charge will flow until the potential becomes the same.
$$V_1=V_2$$
$$\Rightarrow \dfrac{Q_1}{C_1}=\dfrac{Q_2}{C_2}$$    
$$\Rightarrow Q_1= Q_2$$                                 $$[\because C_1= C_2=C]$$
Applying charge conservation: $$Q_1+ Q_2= Q$$
     
            $$\therefore Q_1 = Q_2 = \dfrac{Q}{2}$$ 

$$\textbf{Step 3: Calculate Potential on each capacitor after they are connected }$$
               $$V_1 = \dfrac{Q_1}{C} = \dfrac{Q}{2C} , \ V_2 = \dfrac{Q_2}{C} = \dfrac{Q}{2C}$$

               $$\Rightarrow V_1 = \dfrac{V}{2}, \ V_2 = \dfrac{V}{2}$$

$$\textbf{Step 4: Calcula final energy stored in combined system }$$
                $$U_f = \dfrac{1}{2} C_1V_1^2 + \dfrac{1}{2} C V_2^2$$

                     $$= \dfrac{1}{2} C \left(\dfrac{V}{2}\right)^2 + \dfrac{1}{2} C \left(\dfrac{V}{2}\right)^2$$

                $$\Rightarrow U_f = \dfrac{CV^2}{4}$$                              $$....(2)$$

$$\textbf{Step 5: Calculate Ratio of energies }$$
From equation $$(1)$$ and $$(2)$$
               $$\dfrac{U_f}{U_i} = \dfrac{CV^2/4}{CV^2/2}$$

               $$\Rightarrow \dfrac{V_f}{V_i} = \dfrac{1}{2}$$


Electrostatic energy of $$3.5 \times {10}^{-4}J$$ is stored in a capacitor at $$700 V$$. What is the charge on the capacitor ? 

Given : $$u=3.5\times { 10 }^{ -4 }J$$, $$V=700V$$
$$u=\dfrac { 1 }{ 2 } QV$$
$$Q=\dfrac { 2u }{ V }$$
   $$ =\dfrac { 2\times 3.5\times { 10 }^{ -4 } }{ 700 }$$
$$ Q={ 10 }^{ -6 }C$$


(a)  Obtain a relation for equivalent capacitance of the series combination of capacitors. Draw a circuit diagram.
(b)  $$10$$ capacitors each of capacity $$10\mu F$$ are joined first in series and then in parallel. Write the value of product of equivalent capacitances.
(c)  What will be the value of capacitance of a $$4\mu F$$ capacitor if a dielectric of dielectric constant $$2$$ is inserted fully between the plates of parallel plate capacitor.

(a)
Capacitors in series connected to a voltage source is shown in the attached figure.
Charge across capacitors is equal. Hence, $$Q = Q_1 = Q_2$$
By KVL, $$V = V_1 + V_2$$
From definition of capacitance, $$Q=CV, \quad Q_1 = Q = C_1 V_1, \quad Q_2 =Q= C_2 V_2$$ 
where $$C$$ is equivalent capacitance
From above equations: 
$$\cfrac{Q}{C} =\cfrac{Q}{C_1} + \cfrac{Q}{C_2}$$
$$\cfrac{1}{C} = \cfrac{1}{C_1} + \cfrac{1}{C_2}$$

(b)
For capacitors is series, equivalent capacitance is: 
$$\cfrac{1}{C_s} = \cfrac{1}{C_1} + \cfrac{1}{C_2} + .....+\cfrac{1}{C_{10}}$$
$$\cfrac{1}{C_s} = \cfrac{1}{10} + \cfrac{1}{10} + .... 10\ times$$
$$C_s = 1\ \mu F$$
For capacitors is parallel, equivalent capacitance is: 
$$C_p = C_1 + C_2 + ..... + C_{10}$$
      $$ = 10 + 10 + ......10\ times$$
      $$= 100 \ \mu F$$

Product of equivalent capacitance is $$C_s C_p = 100 \times 10^{-6} \times 1 \times 10^{-6} = 10^{-10} F^2$$ 

(c)
Capacitor of capacitance with a dielectric is:
$$C = \cfrac{K \varepsilon_o A}{d}$$
    $$= 2 \times 4 = 8\ \mu F$$

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Define an equipotential surface,. Draw equipotential surfaces: In a constant electric field in Z - direction Why the equipotential surface about a single charge are not equidistant?

An equipotential surface is the surface with a constant value of potential at all points on the surface. Equipotential surface :
In case of electric field in Z -direction
Potential of a point charge at a distance $$r = \dfrac{1}{4\pi \varepsilon _{0}} \dfrac{q}{r}$$
$$\therefore V \alpha \dfrac{1}{r}$$
Hence equipotential surface about a single chrge are not equidistant.
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Define equipotential surface.

Any surface over which the potential is constant is called an equipotential surface.In other words, the potential difference between any two points on an equipotential surface is zero.

Some important properties of equipotential surfaces :
1. Work done in moving a charge over an equipotential surface is zero.
2. The electric field is always perpendicular to an equipotential surface.
3. The spacing between equipotential surfaces enables us to identify regions of strong and weak fields.
4. Two equipotential surfaces can never intersect. If two equipotential surfaces could intersect, then at the point of intersection there would be two values of electric potential which is not possible.





Derive the expression for potential energy of a system of two charges in the absence of the external electric field.

Coulombic force acting on a test charge $$F = \dfrac{k Q q}{r^2}$$ where $$k = \dfrac{1}{4\pi \epsilon_o}$$
Potential energy of the two charges $$U = -\int F . dr$$
$$\therefore$$ $$U = -\int \dfrac{kQq}{r^2} dr$$
Or $$U = -kQq\int \dfrac{1}{r^2} dr$$
Or $$U = -kQq\times \dfrac{-1}{r} $$
$$\implies$$ $$U = \dfrac{kQq}{r}$$


Three capacitors of capacitances $$4 \mu F, 6 \mu F$$ and $$8\mu F$$ are connected in parallel.
(i) What is the ratio of charges?
(ii) What is the ratio of potential differnces?

(ii) : Potential difference is same across the electrical components which are connected in parallel to each other.
Thus potential difference across the three capacitors is same (or equal), hence ratio of potential differences is  $$1:1:1$$.
(i) : Given: $$C_1 = 4\mu F$$           $$C_2 = 6\mu F$$           $$C_3 = 8\mu F$$
Charge on capacitor, $$Q = CV$$
$$\implies$$  $$Q \propto C$$           (as $$V$$ is same across the capacitors)

Thus ratio of charges  $$Q_1:Q_2:Q_3 = C_1:C_2:C_3$$
$$\therefore$$   $$Q_1:Q_2:Q_3 = 4:6:8$$
$$\implies$$  $$Q_1:Q_2:Q_3 = 2:3:4$$


Derive an expression for energy stored in a capacitor. In which form energy is stored?

Let us consider a capacitor of capacitance C and potential difference V between the plates.

Let the charge on one plate be +q and -q on the other.

Suppose the capacitor is being charged gradually.

Now,at any stage the charge on capacitor is q.

Therefore, the potential difference = $$ \frac{q}{C} $$

Small amount of work doe in giving n additional charge dq to the capacitor is
dW= $$\dfrac{q}{C}*dq$$

Total work done in giving a charge Q to the capacitor  is
W=$$\int dW$$

W= $$ \int_{0}^{Q} \frac{Q}{C}dq $$

 W = $$\dfrac{Q^{2}}{C}$$

Energy = E
E= $$\dfrac{Q^2}{2C}$$ = $$\dfrac {CV^2}{2}$$ = $$\dfrac{QV}{2}$$

The energy is stored in the form of potential energy.


A capacitor of capacity $$C$$ is charged to a potential $$V$$. Then it is taken out and connected in parallel with an uncharged identical capacitor. Then charge on each capacitor will be _____.

Case $$I:$$$$\quad$$ $$Q=CV$$
Case $$II:$$ $$Q=2CV$$
$$Q=q_1+q_2$$
so charges on each capacitor $$=CV$$

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Increase or decrease of electric potential energy depends on the distance or on the high of the two change?

Electric potential is the amount of potential energy per unit charge. If you pick a distance r from a point charge, $$q_1$$, the potential will be;

$$V_R=k\times \dfrac{q_1}{r}$$

$$k$$ is Coulomb's constant. When you place a second charge, $$q_2$$, at r, the potential energy, $$W$$, will be;

$$W=V(r)q_2$$.

So potential is the scaling factor for the potential energy. Therefore, it increases as you get closer to the charge source.



A $$10\mu F$$ capacitor is connected to a $$100V$$ battery. What is the electrostatic energy stored?

$$U=\dfrac{1}{2}CV^2$$
$$U=\dfrac{1}{2}\times 10\times 10^{-4}\times 250^2$$
$$U=5\times 10^{-4}\times 250^2$$
$$U=31.25J$$


A capacitor is connected to a $$12\ V$$ battery through a resistance of $$10\Omega$$. it is found that the potential difference across the capacitor rises to $$4.0\ V$$ in $$1\ \mu s$$. Find the capacitance of the capacitor.
(GIven $$\ell n 3=1.0986, \ell$$ n2 = 0..693)


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Derive the expression for capacitance of the spherical conductor


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Define an equipotential surface. Draw equipotential surfaces: In the case of a single point charge 

An equipotential surface is the surface with a constant value of potential at all points on the surface. Equipotential surface :
In case of a single point charge
Hence point charge is positive, if it is negative then electric field will be radially in ward but equipotential surfaces are same and same and are concentric spheres with centres at the charge.
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Derive an expression for effective capacitance of three capacitors connected in parallel.
1113122_001eeb2da6404cc585c6e8bddf45ff97.png

In the given figure, capacitors $${{C}_{1}},{{C}_{2}}\,and\,\,{{C}_{3}}$$are connected in parallel. The same potential is applied across all the three capacitors. Let $${{Q}_{1}},{{Q}_{2}}\,and\,{{Q}_{3}}$$are charges on the three plates of capacitors such that

$$ {{Q}_{1}}={{C}_{1}}V\, $$

$$ {{Q}_{2}}={{C}_{2}}V $$

$$ {{Q}_{3}}={{C}_{3}}V $$

The total charge stored is $$Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}$$ and equivalent capacitance is $${{C}_{eq}}=\dfrac{Q}{V}$$

So,

$$ {{C}_{eq}}=\dfrac{{{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}}{V} $$

$$ =\dfrac{{{Q}_{1}}}{V}+\dfrac{{{Q}_{2}}}{V}+\dfrac{{{Q}_{3}}}{V} $$

$$ {{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}} $$



Two identical non-conducting spherical shells having equal charge Q, which is uniformly distributed on it, are placed at a distance d apart, from where they are released. Find out kinetic energy of each sphere when they are at a large distance.

Formula,

$$F=\dfrac{1}{4\pi \varepsilon _0}\dfrac{Q^2}{d^2}$$

$$KE=\int_{d}^{r}F dr=\int_{d}^{r}\dfrac{1}{4\pi \varepsilon _0}\dfrac{Q^2}{d^2}dr=\dfrac{Q^2}{4\pi \varepsilon _0}\left [ -\dfrac{1}{r} \right ]_d^r$$

$$\therefore KE=\dfrac{Q^2}{4\pi \varepsilon _0}\left [ \dfrac{1}{d}-\dfrac{1}{r} \right ]$$

where, r is the maximum distance of identical spherical charges from each other


A parallel-plate capacitor with the plate area $$100 \,cm^2$$ and the separation between the plats $$1.0 \, cm$$ is connected across a battery of emf $$24$$ volts. Find the force of attraction between the plates.

The surface area of the capacitor plates $$A=100cm^2=10^{}m^2$$
The separation between the plates is $$d=1cm=10^{-2}m$$
The emf of the battery is $$V=24V_0$$

$$\therefore$$ The capacitance od the capacitor is given as:
$$C=\dfrac {\epsilon_0A  }{ d }$$

$$=\dfrac { 8.85\times 10^{-12}\times 10^{-2} }{10^{-2}  }$$

$$=8.85\times 10^{-12}$$

$$\therefore$$ The energy stored in the capacitor is given as:
 $$E=(1/2)CV^2$$

$$=(1/2)\times 10^{-12}\times (24)^2=2548.8\times 10^{-12}$$

$$\therefore$$ The forced attraction between the plates $$=\dfrac { E}{  d} =\dfrac {2548.8\times 10^{-12}  }{10^{-2}  }= 2.54\times 10^{-7}N$$


In a hydrogen atom, the electron and proton and bound at a distance of about 0.53 A.
Estimate the potential energy of the system in eV, taking the zero of potential energy at infinite separation of electron from proton.

$$r=0.53\ A$$
$$=0.53 \times 10^{10}\ m$$
Charge of electron $$-e$$
Charge of proton $$=e$$
$$\therefore $$ Potential energy $$=\dfrac{1}{4 \pi \varepsilon_0}.\dfrac{e(-e)}{r}=\dfrac{-e^2}{4 \pi \varepsilon_0 r}$$
$$V=\dfrac{-(1.6)^2 \times 10^{-38} \times 9 \times 10^9}{0.53 \times 10^{-10}}$$
$$=\dfrac{-(1.6)^2 \times 10^{-38} \times 9 \times 10^9}{0.53 \times 10^{-10} \times 1.6 \times 10^{-19}} eV$$
$$=\dfrac{-1.6 \times 9}{0.53}eV=-27.2 eV$$


Find the equivalent capacitance of the infinite ladder shown in the figure between the point $$A$$ and $$B$$.
1552051_eec7b58d2db24ad6bfafd2aaf855d302.png

Let the equivalent capacitance be C. Since it is an infinite series. So, there will be negligible change if the arrangement is done an in fig II

The $$2\mu F$$ and $$C$$ are in series with each other and they are in parallel to $$1\mu F$$ capacitor.
$$C_{eq} = \dfrac{2 \times C}{2 + C} + 1 $$

$$\Rightarrow C = \dfrac{2C + 2 + C}{2 + C}$$

$$\Rightarrow (2 +C) \times C = 3 C + 2$$
$$\Rightarrow C^2 - C - 2 = 0$$
$$\Rightarrow (C - 2) (C + 1) = 0$$
$$C = - 1$$ (Impossible)

So, $$C = 2 \mu  F$$

1545039_1552051_ans_bf1a1d2c0404434c8a662bb418767395.png


In a hydrogen atom, the electron and proton and bound at a distance of about 0.53 A.
 What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation?

At $$1.06\ A^o$$ distance, $$PE=-13.6eV$$
It is brought to zero by adding $$+13.6eV$$
$$\therefore PE$$ at $$0.53$$ is $$-13.6eV$$ and energy needed is $$13.6eV$$


Two point charge $$q_1$$ and $$q_2$$ are kept r distance apart in a uniform external electric field $$\vec{E}$$. Find the amount of work done in assembling this system of charges.

Let the charge $$q_1$$ travels $$r_1$$ distance and $$q_1$$ travels
The work done in bringing the charge $$q_1$$ in the field $$to$$
$$W_1 = F_1 \times r_1$$
      $$=q_1E \times r_1$$
the work done in bringing the second charge $$\to$$
$$W_2 = F_2 \times r_2$$
      $$= q_2 E \times r_2$$
and the work is also done to overcome the force of the charge on one-another.
$$W_3 = \dfrac{1}{4\pi\in_0}. \dfrac{q_1q_2}{r^2}$$
So, total work $$= q_1Er_1 + q_2Er_2 +\dfrac{1}{4\pi \in_0} \dfrac{q_1q_2}{r^2}.$$ 


In the figure given below, Find the
total energy stored in the network.
1700251_f43f7bd6248a428baf7b7bf92ff20dcf.png

Given circuit,
$$(C_1 = C_5 = 4 \mu F, C_2 = C_3 = C_4 = 2 \mu F)$$
Ref. image

From about circuit diagram $$C_1 $$ & $$C_2$$ and $$C_4$$ & $$C_5$$ are shorted with wire. So total potential drop to move from $$C_1 $$ to $$C_2 $$ and $$C_4 $$ to $$C_5$$ will be zero

$$\therefore C_{eq} = C_3 = 2 \mu F$$

Charge supply by battery

$$q = C_{eq} \times V$$

$$= 2 \times 10^{-6} \times 5$$

$$q = 10 \times 10^{-6}$$

$$q = 10 \mu C$$

total energy stored 

$$U = \dfrac{1}{2} (q v) = \dfrac{1}{2} \times 10 \times 5 = 25 \mu J$$

1497131_1700251_ans_301d34fcd376478ba1976c7b00fe5b72.png


Obtain the formula for stored energy in a charged capacitor.

Let at any  instant t, the charge $$q^{1}$$ be already transferred from one plate to another. In this instant, the potential difference $$V^{1}$$ between the plates of a capacitor,
$$V^{1} = \dfrac{q^{1}}{C}$$
Now the work done in transferring the charge $$dq$$ ,
$$dW = V^{1}dq$$ ........ .........(1)
This work is stored in the form of potential energy in the capacitor. Total work done in charging the capacitor from zero to charge Q
$$W = \displaystyle \int_{0}^{Q} V^{1}dq$$
or,  $$W = \displaystyle \int_{0}^{Q} \dfrac{q^{1}}{C} dq$$
or,  $$W = \dfrac{1}{C} \displaystyle \int_{0}^{Q} q^{1}dq$$     
or,  $$W = \dfrac{1}{C} \left [ \dfrac{q^{1^{2}}}{2} \right ]^Q_0$$
$$\Rightarrow$$    $$W = \dfrac{1}{2} \dfrac{Q^2}{C}$$
Thus, the energy stored in charged capacitor,
$$U = W $$
$$U = \dfrac{1}{2} \dfrac{Q^2}{C}$$               ....(2)
Putting $$Q = CV$$ in eqn. (2)
$$U = \dfrac{1}{2} CV^2$$         ..(3)
Now, putting $$\dfrac{Q}{C} = V$$ in eqn. (2)
$$U = \dfrac{1}{2} QV$$           ..(4)
Thus, energy stored in charged capacitor
$$U = \frac{1}{2} \dfrac{Q^2}{C}$$                       ..(5)
$$ = \frac{1}{2} CV^2 = \dfrac{1}{2} QV$$
This formula is electric potential energy of a charged conductor. Consider two capacitors 1 and 2 whose area A is same. The capacitance of capacitor 1 is half of that of capacitor 2. Let the charges on both the capacitors be q, then the electric field between the two plates , E $$= \dfrac{q}{\varepsilon_0 A}$$ will be same. Now, due to difference in capacitance, it is clear from equation (2), that the potential energy of capacitor 1 is double the capacitor.



Define electric capacitance and write its S.I.unit.

The ratio of charge given to a conductor and potential raised due to this charge is called electric capacitance of the conductor.
From equation : 
C $$= \frac{q}{V}$$
Thus , the electric capacitance of a conductor is a constant. Its  value depends upon the nature and area of the conductor and on the environment around the conductor and also on the other conductors placed near the conductor.
Again from $$C$$ $$= \frac{q}{V}$$ , if $$V$$ =1 volt, then $$C = q $$
i.e. the electrical capacitance of a conductor of a conductor is equal to charge of conductor when the applied potential on a conductor take as unity.
Unit and Dimensions:
$$\because C = \frac{q}{V}$$
$$\therefore$$ Unit of capacitance $$= \frac{coulomb}{volt} =$$ CV$$^{-1} =$$ Farad (F)
$$\because$$ 1CV $$^{-1} =$$ 1F
If q = 1C, V = 1 volt, then C = 1F
i.e., 'If on giving a charge of 1 C to a conductor its potential is increased by 1 volt, then the capacitance of the conductor will be 1F'.
Farad is a big unit of capacitance. In practice other small units e.g, $$\mu$$F and $$\mu \mu F$$ or pF are used.
$$1 \mu$$F $$= 10^{-6}$$F
and $$1 \mu \mu$$F = 1 pF = $$10^{-12}$$F
Dimensional formula:
$$\because$$   $$C = \frac{q}{V} = \frac{q}{\frac{W}{q}} = \frac{q^2}{W} = \frac{i^2 t^2}{W}$$
$$\therefore$$ Dimensional formula of C $$= \frac{[A^2 T^2]}{M^1 L^2 T^{-2}} $$
                                               $$= [M^{-1} L^{-2} T^{4} A^2]$$ 
Graph : If a graph is plotted between $$q$$ and $$V$$ , then the straight line is obtained.
The gradient of the graphical line $$= \tan \theta = \frac{q}{V} = C$$
i.e., the gradient of the graph provides the capacitance of the conductor.

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In which from the energy of charged capacitor is stored ?

Energy of the capacitor is stored as $$electric\ energy$$ between the plates of capacitor .


Conceptual Questions
The sum of the charges on both plates of a capacitor is zero. What does a capacitor store?


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Match the entries of Column I with the entries of Column II

Generally the charge inside hollow conductor is zero and filed inside it also zero.
For A:  Due to q, charge will be induced on the conductor, such that net field

due to q and induced charge becomes zero at any point inside the

conductor. Since $$E=0$$ everywhere inside the conductor, so potential is

constant inside and same as that of surface of conductor.
thus, $$A - 1,4$$
For B:  As charge q inside the conductor so both field and potential inside the conductor will vary. thus, $$B - 3,5$$
For C: Due to $$q_2$$, field and potential both will vary inside because inside charge system has nothing to do with outside system.
thus, $$C - 3,5$$




The plates of a parallel plate capacitor have an area of $$90\ cm^2$$ each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

$$\textbf{Step 1: Capacitance of the capacitor(C)}$$
Capacitance of parallel plate capacitor is given by:
               $$C = \dfrac{\in_0A}{d}$$                                                                   $$  ....(1)$$

$$\textbf{Step 2: Electrostatic energy stored by the capacitor(U)}$$

                $$U = \dfrac{1}{2} CV^2$$$$ = \dfrac{1}{2} \dfrac{A \in_0}{d} V^2$$                                       $$....(2)$$

                    $$= \dfrac{1}{2} \times \dfrac{90\times 10^{-4}\ m^2 \times 8.85 \times 10^{-12} \times (400V)^2}{(2.5\times 10^{-3}\ m)}$$
                     $$ = 2.55\times 10^{-6} J$$

$$\textbf{Step 3: Energy per unit volume(u)}$$
                 
Energy per unit volume :  $$u = \dfrac{U}{V'}= \dfrac{U}{Ad}$$                                $$....(3)$$

where $$V' = $$ Volume of the capacitor

$$\therefore u= \dfrac{2.55\times 10^{-6}J}{90\times 10^{-4}\ m^2\times2.5\times 10^{-3}\ m} = 0.113 J m^{-3}$$

$$\textbf{Step 4: Relation between u and electric field (E)}$$
From eq $$(2)$$ and $$(3)$$
                     $$u =\dfrac{U}{V'}$$$$ = \dfrac{\dfrac{1}{2} \dfrac{A \in_0}{d} V^2}{Ad}$$

                 $$\Rightarrow u = \dfrac{1}{2} \in_0 \left(\dfrac{V}{d}\right)^2$$

$$\dfrac{V}{d} = $$ Electric field ntensity $$=E$$

$$\therefore u = \dfrac{1}{2} \in_0 E^2$$


Two uniformly large parallel thin plates having charge densities $$+\sigma$$ and $$-\sigma$$ are kept in the XZ-plane at 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?

Given two uniformly large parallel plate having charge densities $$+\sigma$$ and $$-\sigma$$ kept in x-z plane at distance $$d$$ apart.
We have to sketch an equipotential surface due to electric field between the plates.
Here electric field $$E$$ between the plates is given by $$\dfrac{\sigma}{\varepsilon _{0}}$$. Now, $$V=Ed=\dfrac{\sigma}{\varepsilon _{0}d}$$
So, any set of equidistant points between the plates will have same electric potential. 
We also have electric field between the plates if a particle of mass $$m$$ and charge $$-q$$ remains stationary between the plates.
So, the gravitational force experienced by charge $$=mg$$
Electrostatic force experienced by the charge $$=qE=\dfrac{q\sigma}{\varepsilon _{0}}$$
Now, as the charge is static, 
$$mg=qE$$
$$\Rightarrow \: mg=\dfrac{q\sigma}{\varepsilon _{0}}$$
$$\Rightarrow \: \sigma =\dfrac{mg\varepsilon _{0}}{q}$$
Thus, $$E=\dfrac{\sigma}{\varepsilon _{0}}=\dfrac{mg\varepsilon _{0}}{\varepsilon _{0}q}=\dfrac{mg}{q}$$
The direction of electric field is downward i.e, along $$-y$$ axis.

The equipotential surface is as shown:



Find the equivalent capacitances of the combinations shown in figure between the indicated points.
1721358_f64ec98419d042869e65d17abd182895.png

From Above figure,
a) $$C_{ef} = \dfrac{4}{3} + \dfrac{8}{3} + 4 = 8 \mu F$$

b) $$C_{ef} = \dfrac{3}{8} + \dfrac{32}{12} + \dfrac{32}{12} + \dfrac{8}{6} = \dfrac{16 + 32}{6} = 8 \mu f$$

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Each capacitance shown in the figure is in $$\mu F$$. Find the charge on $$6\mu F$$ in $$\mu C$$.
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Three capacitors C$$_1$$, C$$_2$$ and C$$_3$$ are connected in parallel combination. Obtain the relation for equivalent capacitance .

Parallel Combination of Capacitors : 
This combination is adopted when greater capacity is required from smaller capacities. In this combination one plate of each capacitor is connected at one junction and other plates on other junction as shown in the diagram. the junctions are A and B. Suppose +q charge is given to this combination. This charge is distributed on capacitors according to their capacities.
$$\therefore$$   $$q_1$$ + $$q_2$$ + $$q_3$$ ...........(1)
Since all the capacitors are joined between A and EI therefore the potential difference of all the capacitor will be same.
$$\therefore$$ V $$= \frac{q_1}{C_1} = \frac{q_2}{C_2} = \frac{q_3}{C_3}$$
$$\therefore$$ $$q_1 =$$$$C_1V$$, $$q_2 =$$$$C_2V$$, $$q_3 =$$$$C_3V$$
If $$C$$ be the equivalent capacity combination,
then $$q = CV $$
Therefore from equation (1),
$$CV = C_1V + C_2V + C_3V$$
Similarly $$n$$ capacitors can be joining in parallel :
$$C$$ = $$C_1$$ + $$C_2$$ + $$C_4$$ + .... + $$C_n$$


The liquid-drop model of the atomic nucleus suggests high-energy oscillations of certain nuclei can split the nucleus into two unequal fragments plus a few neutrons. The fission products acquire kinetic energy from their mutual Coulomb repulsion. Assume the charge is distributed uniformly throughout the volume of each spherical fragment and, immediately before separating, each fragment is at rest and their surfaces are in contact. The electrons surrounding the nucleus can be ignored. Calculate the electric potential energy (in electron volts) of two spherical fragments from a uranium nucleus having the following charges and radii: $$38e$$ and $$5.50 \times 10^{-15} \,m$$, and $$54e$$ and $$6.20 \times 10^{-15} \,m$$.


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Calculating Capacitance
A $$50.0-m$$ length of coaxial cable has an inner conductor that has a diameter of $$2.58 \,mm$$ and carries a charge of $$8.10 \mu C$$. The surrounding conductor has an inner diameter of $$7.27 \,mm$$ and a charge of $$8.10 \mu C$$. Assume the region between the conductors is air.
(a) What is the capacitance of this cable?
(b) What is the potential difference between the two conductors?


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Conceptual Questions
Describe the equipotential surfaces for (a) an infinite line of charge and (b) a uniformly charged sphere.


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A solid sphere of radius $$R$$ has a uniform charge density $$\rho$$ and total charge $$Q$$. Derive an expression for its totalelectric potential energy. Suggestion: Imagine the sphere is constructed by adding successive layers of concentric shells of charge $$dq =(4\pi r^2 \,dr)\rho$$ and use $$dU =V dq$$


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Calculating Capacitance
(a) Regarding the Earth and a cloud layer $$800 \,m$$ above the Earth as the "plates" of a capacitor, calculate the capacitance of the Earthcloud layer system.Assume the cloud layer has an area of $$1.00 km^2$$ and the air between the cloud and the ground is pure and dry. Assume charge builds up on the cloud and on the ground until a uniform electric field of $$3.00 \times 10^6 \,N/C$$ throughout the space between them makes the air break down and conduct electricity as a lightning bolt.
(b) What is the maximum charge the cloud can hold?


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Class 12 Engineering Physics Extra Questions