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CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 7 - MCQExams.com
CBSE
Class 12 Medical Physics
Electrostatic Potential And Capacitance
Quiz 7
The dimensional formula of electric potential is given by
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$$[ML^2T^{-3} A^{-1}]$$
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$$[ML^2T^{-2} A^{-1}]$$
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$$[ML^2T^{-1} A^{-1}]$$
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$$[ML^2T^{-2}]$$
Explanation
Electric potential is defined as the work done to move per unit positive charge from one point to another point.
The SI unit of electric potential is $$volt(V)$$.
$$V=\dfrac{W}{Q}$$
Hence dimension is $$\dfrac{[ML^2T^{-2}]}{[AT]}$$
$$=[ML^2T^{-3}A^{-1}]$$
Answer-(A)
A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one
complete revolution is_______
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zero
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positive
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negative
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zero if the charge Q is at the center and nonzero otherwise.
Explanation
The net displacement round one complete circle is 0.
So, the work done is 0.
The amount of energy that a unitary point electric charge would have, if located at any point in space, is defined its:
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electric potential energy
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electric potential
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electric potential difference
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electric field
Explanation
The electric potential is the electric potential energy per unit charge.
A charge $$q$$ is placed at $$A(2,3,3)$$ in the $$XYZ$$ co-ordinate system. Find the electric potential at $$B(-2,3,3)$$
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$$\dfrac{q}{4 \pi \epsilon_o}$$
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$$\dfrac{q}{16 \pi \epsilon_o}$$
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$$\dfrac{3q}{16 \pi \epsilon_o}$$
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None of these
Explanation
distance=$$r=\sqrt{2+2)^2+(3-3)^2+(3-3)^2}=4$$
$$V=\dfrac{q}{4\pi\epsilon_o r}$$
$$\implies V=\dfrac{q}{16\pi\epsilon_o}$$
Answer-(B)
How many 6m
F, 200 V condensers are needed to make a condenser of 18 mF, 600 V?
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9
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18
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3
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27
Explanation
$$6mF,200V$$
To get potential of 600V, 3 capacitors (6mF,200V) must be connected in series
$$C=6mF\\ { \left( { C }_{ eq } \right) }^{ -1 }=\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 6 } \Rightarrow { \left( { C }_{ eq } \right) }^{ -1 }={ 2 }^{ -1 }\\ \Rightarrow { C }_{ eq }=2mF$$
To get 18mF, the above must be repeated combination for 9 times which should be connected in parallel, so that we get $$C_{eq\; final}=18mF$$
So, finally we would get 18mF,600V condenser.
So, total number of required condenser = $$9\times 3=27$$
A uniform electric field of magnitude $$100V/m$$ in space is directed along the line $$y=3+x$$. Find the electric potential difference between $$(1,3)$$ and $$(3,1)$$.
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$$100V$$
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$$200\sqrt2 V$$
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$$200V$$
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$$0$$
Explanation
Given electric field, $$E=100\, V/m$$ directed along line $$y=3+x$$
We have to find the electric potential difference between $$(1, 3)$$ and $$(3, 1)$$.
We know that the potential difference $$V$$ is related to electric field $$E$$ by the relation, $$E=\dfrac{-dV}{dr}$$
or, $$dV=-Edr$$
So, potential difference between the points $$(3, 1)$$ and $$(1, 3)$$ is:
$$dV=-100(1-3)-100(3-1)=0\, V$$
Identify the true expression:
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$$V=JC^{-1}$$
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$$V=JC$$
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$$V=JC^{-2}$$
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$$V=JC^{-3}$$
Explanation
Electric potential is defined as the work done to move a unit positive charge from one point to another point.
The SI unit of electric potential is $$volt(V)$$.
Hence, $$V=\dfrac{J}{C}$$
Hence, $$V=\dfrac{J}{C}=JC^{-1}$$
Answer-(A)
Electric potential at any point is $$V = -5x +3y +\sqrt{15}z$$, then the magnitude of electric field is __________ $$N/C$$
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$$3\sqrt2$$
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$$4\sqrt2$$
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$$7$$
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$$5\sqrt2$$
Explanation
$$V=-5x+3y+\sqrt{15}z$$
$$\vec{E}=-\dfrac{\partial V}{\partial x}\hat{i}-\dfrac{\partial V}{\partial y}\hat {j}-\dfrac{\partial V}{\partial z}\hat{k}$$
Now $$\dfrac{\partial V}{\partial x}=-5$$
$$\dfrac{\partial V}{\partial y}=3$$
$$\dfrac{\partial V}{\partial z}=\sqrt{15}$$
$$\vec{E}=5\hat{i}-3\hat{j}-\sqrt{15}\hat{k}$$
$$\implies E=\sqrt{5^2+(-3)^2+(\sqrt{15})^2}$$
$$\implies E=\sqrt{25+9+15}$$
$$\implies E=7 N/C$$
Answer-(C)
In the definition of electric potential, the electric potential at infinity is assumed to be
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infinity
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zero
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$$1$$
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None of these
Explanation
Potential $$(V)$$ at a point varies inversely with the distance $$r$$ i.e, $$V\propto \dfrac{1}{r}$$; when $$r=\infty$$
$$V=0$$
So, the electric potential at infinity is assumed to be zero.
Some charge is being given to a conductor. Then its potential:
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is maximum somewhere between surface and centre.
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is maximum at surface.
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is maximum at centre.
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remains same throughout the conductor.
Explanation
Given that some charge is given to a conductor then the whole charge is distributed over its surface only. Inside of conductor, electric field is zero whereas potential is same as on the surface. Hence, throughout the conductor, potential is same i.e, the whole conductor is equipotential.
Electric potential can be calculated in a
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static electric field
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dynamic electric field
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both static and dynamic electric field
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neither static nor dynamic electric field
Explanation
An electric potential can be calculated in either a static (or time invariant) or a dynamic (varying with time) electric field at a specific time in units of Joules per Coulomb or Volts.
If the plates of a parallel plate charged capacitor are not parallel, the interface charge density is
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is higher at the closer end
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is non-uniform
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Is higher at inclined plate.
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Is uniform
Calculate the potential at a point due to a charge of $$4\times { 10 }^{ -7 }C$$ located at $$0.09 m$$ away from it.
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$$4\times { 10 }^{- 4 }V$$
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$$4\times { 10 }^{ 4 }V$$
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$$7\times { 10 }^{ 4 }V$$
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$$4\times { 10 }^{ 2 }V$$
Explanation
Given,
$${ q }_{ 1 }=4\times { 10 }^{ -7 }C$$, $$r=0.09m$$
Therefore, the potential due to the charge $${ q }_{ 1 }$$ at the point is
$$V=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { { q }_{ 1 } }{ r }$$
$$ =\dfrac { 9\times { 10 }^{ 9 }\times 4\times { 10 }^{ -7 } }{ 0.09 } =4\times { 10 }^{ 4 }V$$
Point A is at a lower electrical potential than point B. An electron between them on the line joining them will
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move towards A
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move towards B
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move at right angles to the line joining A and B
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remain at rest
Explanation
Given that point $$A$$ is at lower electric potential than point $$B$$. The electron between them on line joining will move.
We have to find where this electron moves.
Since we know that electric currents move from a higher potential or a lower potential. Also, electrons move in the direction opposite to electric current. So the electron on the line joining two points $$A$$ and $$B$$ will move from lower to higher potential i.e, it will move towards $$B$$.
The capacitance of a parallel plate capacitor with air as dielectric is C. If a slab of dielectric constant K and of the same thickness as the separation between the plates is introduced so as to fill 1/4th of the capacitor (shown in figure), then the new capacitance is
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$$(K+2)\dfrac{C}{4}$$
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$$(K+3)\dfrac{C}{4}$$
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$$(K+1)\dfrac{C}{4}$$
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None of these
Explanation
Given,
Capacitance of a parallel plate capacitor with air as dielectric is $$=C$$
Dielectric constant of the slab$$=K$$
Sepration between the plates is $$=\dfrac{1}{4}^{th}$$
Capacitance, $$C=\dfrac{{\epsilon}_0A}{d}$$
As one-fourth of capacitor is filled with dielectric of constant K, then,
$$C_1=\dfrac{K{\epsilon}_0A/4}{d}$$
and $$C_2=\dfrac{{\epsilon}3A/4}{d}$$
Both $$C_1$$ and $$C_2$$ are in parallel.
$$\therefore C_p=C_1+C_2=\dfrac{K{\epsilon}_0A}{4d}+\dfrac{3{\epsilon}_0A}{4d}$$
$$=(K+3)\dfrac{{\epsilon}_0A}{4d}=(K+3)\dfrac{C}{4}$$
So, The new capacitance $$=(K+3)\dfrac{C}{4}$$.
Two point charges, each of charge $$q$$, are placed at a separation of $$2a$$. The electric potential at their midpoint will be :
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Zero
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$$\dfrac{q}{2 \pi {\varepsilon}_{0} a}$$
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$$\dfrac{q}{8 \pi {\varepsilon}_{0} a}$$
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$$\dfrac{q}{2 \pi {\varepsilon}_{0} {a}^{2}}$$
Explanation
Potential at any point is given by $$V = \dfrac{q}{4\pi \epsilon_o d}$$
where $$d$$ is the distance of that point from the charge.
Thus potential at the midpoint P $$V_p = \dfrac{q}{4\pi \epsilon_o a} + \dfrac{q}{4\pi \epsilon_o a}$$
$$\therefore$$
$$V_p = \dfrac{q}{2\pi \epsilon_o a}$$
The potential difference between the two plates of a parallel plate capacitor is constant. When air between the plates is replaced by dielectric material, the electric field intensity :
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Decreases
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Remains unchanged
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Becomes zero
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Increases
Explanation
In general capacitance of parallel plate capacitor is given by:
$$C=\dfrac{k\epsilon_0 A}{d}$$
where C is capacitance, k is relative permittivity of dielectric material, $$\epsilon_0$$ is permittivity of free space constant, A is area of plates and d is distance between them.
Therefore, the capacitance of parallel plates is increased by the insertion of a dielectric material. Further, the capacitance is inversely proportional to the electric field between the plates, and hence the presence of the dielectric decreases the effective electric field.
Three capacitors of capacitance $$1.0,2.0$$ and $$5.0\mu F$$ are connected in series to a $$10V$$ source. The potential difference across the $$2.0\mu F$$ capacitor is
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$$\cfrac { 100 }{ 17 } $$
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$$\cfrac { 20 }{ 17 } V$$
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$$\cfrac { 50 }{ 17 } V$$
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$$10V$$
Explanation
$${ C }_{ eq }=\cfrac { 10 }{ 17 } \mu F,Q=\cfrac { 100 }{ 17 } \mu C\quad $$
$$\therefore$$ Potential difference across $$2\mu F$$ capacitor
$$=\cfrac { \cfrac { 100 }{ 17 } \mu C }{ 2\mu F } =\cfrac { 50 }{ 17 } V$$
The amount of work done in increasing the voltage across the plates of capacitor from 5V to 10V is 'W'. The work done in increasing it from 10V to 15V will be
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W
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0.6 W
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1.25 W
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1.67 W
Explanation
Let the capacitance of the capacitor be $$C$$.
Energy stored in the capacitor $$E = \dfrac{1}{2}CV^2$$
Work done in changing the potential from 5V to 10V, $$W = \dfrac{1}{2}C (V_2^2 - V_1^2)$$
$$\therefore$$
$$W = \dfrac{1}{2}C (10^2 - 5^2) = \dfrac{1}{2}C (75)$$
Work done in changing the potential from 10V to 15V, $$W' = \dfrac{1}{2}C (15^2 - 10^2)$$
$$\therefore$$
$$W' = \dfrac{1}{2}C (125)$$
$$\implies$$ $$\dfrac{W'}{W} = \dfrac{125}{75} = 1.67$$
We get $$W' = 1.67 W$$
A metal plate of thickness $$d/2$$ is introduced in between the plates of a parallel plate air capacitor with plate separation of $$d$$. Capacity of the metal plate :
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Decreases $$2$$ times
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Increases $$2$$ times
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Remains same
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Becomes zero
Explanation
$$C=\dfrac{\epsilon_oA}{d}$$
when a dielectric is filled ,thickness t
$$C_1=\dfrac{\epsilon_oA}{d-t+\frac{t}{K}}$$
for metal k= infinity ,$$t=\dfrac{d}{2}$$
$$C_1=\dfrac{2\epsilon_oA}{d}=2C$$
The energy stored in a capacitor of capacitance C having a charge Q under a potential $$V$$ is
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$$\dfrac {1}{2}Q^2V$$
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$$\dfrac {1}{2}C^2V$$
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$$\dfrac {1}{2}\dfrac {Q^2}{V}$$
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$$\dfrac {1}{2}QV$$
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$$\dfrac {1}{2}CV$$
Explanation
By energy stored in a charged conductor Suppose, a conductor of capacity C is charged to a potential V and Q the charge on conductor at a instant.
The potential of the conductor
$$V=\frac {Q}{C} $$
Now, work done in bringing a small charge dQ at this potential is,
$$dW = v \, dQ = \frac {Q}{C} dQ$$
$$\therefore$$ Total work done in charging it from 0 to Q is,
$$W= \int_{0}^{Q} dW =\int_{0}^{Q}\frac {Q}{C} dQ$$
$$= \frac {1}{2}\frac {Q^2}{C}$$
This work is stored as the potential energy.
$$U=\frac {1}{2}\frac {Q^2}{C}$$ $$(\because Q=VC)$$
$$U=\frac {1}{2} CV^2 =\frac {1}{2}QV$$
An electron enters into a space between the plates of parallel plate capacitor at an angle of $$\alpha$$ with the plates and leaves at an angle of $$\beta$$ to the plates. The ratio of its $$KE$$ while leaving to entering the capacitor will be :
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$$\left (\dfrac {\cos \alpha}{\cos \beta}\right )^{2}$$
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$$\left (\dfrac {\cos \beta}{\cos \alpha}\right )^{2}$$
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$$\left (\dfrac {\sin \alpha}{\sin \beta}\right )^{2}$$
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$$\left (\dfrac {\sin \beta}{\sin \alpha}\right )^{2}$$
Explanation
We know that Electric field is always perpendicular to the plates of the capacitor.
Let the speed of the electron be $$v_1$$ when it enters the gap between the plates and $$v_2$$ when it leaves the plates.
As the electron only experiences force in the direction perpendicular to the plates, by the conservation of momentum in the direction parallel to the plates, we have
$$v_1\textrm{cos}(\alpha) = v_2\textrm{cos}(\beta) \Rightarrow \dfrac{v_2}{v_1} = \dfrac{\textrm{cos}\alpha}{\textrm{cos}\beta}$$
The ratio of kinetic energy is given by $$(\dfrac{v_2}{v_1})^2 = (\dfrac{\textrm{cos}\alpha}{\textrm{cos}\beta})^2$$
Given below are three schematic graphs of potential energy $$V(r)$$ versus distance $$r$$ for three atomic particles: electron $$(e^{-})$$, proton $$(p^{+})$$ and neutron $$(n)$$, in the presence of a nucleus at the origin $$O$$. The radius of the nucleus is $$r_{0}$$. The scale on the V-axis may not be the same for all figures. The correct pairing of each graph with the corresponding atomic particle is.
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$$(1, n), (2, p^{+}), (3, e^{-})$$
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$$(1, p^{+}), (2, e^{-}), (3, n)$$
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$$(1, e^{-}), (2, p^{+}), (3, n)$$
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$$(1, p^{+}), (2, n), (3, e^{-})$$
Explanation
The interaction forces between the sub-atomic particles are
(i) Electrostatic forces and (ii) Nuclear forces/Strong forces.
The nuclear forces are very large in magnitude compared to electrostatic forces, but they are short distance forces. They only act within the nucleus.
Thus, within the nucleus, the potential is governed by nuclear forces and outside the nucleus, it is governed by the electrostatic forces.
$$\textrm{Case 1: Neutron}$$
Neutron is a chargeless particle. Thus, outside the nucleus, electrostatic potential energy of neutron is zero.
But, within the nucleus, the nuclear forces (attractive) bind the neutron to the nucleus. Hence potential energy is negative.
Thus, Potential energy of neutron corresponds to figure (1)
$$\textrm{Case 2: Proton}$$
proton is a positively charged particle. Thus, outside the nucleus, electrostatic forces on the proton are repulsive. Thus, potential energy of proton is positive.
But, within the nucleus, the nuclear forces (attractive) bind the proton to the nucleus. Hence potential energy is negative.
Thus, Potential energy of proton corresponds to figure (2)
$$\textrm{Case 3: Electron}$$
Electron is a negatively charged particle. Thus, outside the nucleus, electrostatic forces on the electron are attractive. Thus, potential energy of electron is negative.
But, within the nucleus, the nuclear forces and the electrostatic forces are of the same order (since electron is not a intra nuclear particle).
Thus, Potential energy of electron corresponds to figure (3)
Two identical parallel plate capacitance C each are connected in series with a battery of emf, E as shown, If one of the capacitors is now filled with a dielectric of dielectric constant k, the amount of charge which will flow through the battery is $$?$$(neglect internal resistance of the battery)
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$$\displaystyle\frac{k+1}{2(k-1)}CE$$
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$$\displaystyle\frac{k-1}{2(k+1)}CE$$
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$$\displaystyle\frac{k-2}{k+2}CE$$
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$$\displaystyle\frac{k+2}{k-2}CE$$
Explanation
Inital charge on both the capacitor $$Q=\dfrac{CE}{2}$$
final charge$$Q_f=\dfrac{kCE}{K+1}$$
charge flow from batter is $$Q_f-Q=\dfrac{k-1}{2(k+1)}CE$$
A charge $$10nC$$ is situated in a medium of relative permittivity $$10$$. The potential due to this charge at a distance of $$0.1 m$$ is :
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$$900V$$
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$$90V$$
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$$9V$$
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$$0.09V$$
Explanation
Relative permittivity of the charge $$\epsilon_r =10$$
Magnitude of charge $$q = 10nC = 10^{-8} C$$
Potential at a distance $$0.1m$$, $$V_p = \dfrac{q}{4\pi \epsilon d}$$
where $$\epsilon = \epsilon_r \epsilon_o$$
$$\therefore$$
$$V_p = \dfrac{q}{4\pi \epsilon_o \epsilon_r d}$$
Or
$$V_p = \dfrac{9\times 10^9 \times 10^{-8}}{10 \times 0.1} =90 V$$
An insulator plate is passed between the plates of a capacitor. Then, current :
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First flows from A to B and then from B to A
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First flows from B to A and then from A to B
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Always flows from B to A
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Always flows from A to B
Two charges $$+6\mu C$$ and $$-4\mu C$$ are placed $$15 cm$$ apart as shown. At what distances from $$A$$ to its right, the electrostatic potential is zero?
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$$4, 9, 60$$
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$$9, 15, 45$$
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$$20, 30, 40$$
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$$9, 45$$, infinity
Explanation
Let the potential be zero at point $$P$$ at a distance $$x$$ from the charge $$+6\times { 10 }^{ -6 }C$$ at $$A$$ as shown in above figure. Potential at $$P$$
$$V=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { \left( -4\times { 10 }^{ -6 } \right) }{ 15-x } \right]$$
$$0=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { 4\times { 10 }^{ -6 } }{ 15-x } \right]$$
$$0=\dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { 4\times { 10 }^{ -6 } }{ 15-x }$$
$$ \dfrac { 6\times { 10 }^{ -6 } }{ x } =\dfrac { 4\times { 10 }^{ -6 } }{ 15-x }$$
$$\Rightarrow 6\left( 15-x \right) =4x$$
$$\Rightarrow 90-6x=4x$$
$$\Rightarrow 10x=90$$
$$\Rightarrow x=\dfrac { 90 }{ 10 } =9 cm$$
The other possibility is that point of zero potential $$P$$ may lie on $$AB$$ produced at a distance $$x$$ from the charge $$+6\times { 10 }^{ -6 }C$$ at $$A$$ as shown in the figure.
Potential at $$P$$
$$V=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 6\times { 10 }^{ -6 } }{ x } +\dfrac { \left( -4\times { 10 }^{ -6 } \right) }{ \left( 15-x \right) } \right]$$
$$0=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { 4\times { 10 }^{ -6 } }{ \left( x-15 \right) } \right]$$
$$0=\dfrac { 6\times { 10 }^{ -6 } }{ x } -\dfrac { 4\times { 10 }^{ -6 } }{ \left( x-15 \right) }$$
$$\Rightarrow \dfrac { 6\times { 10 }^{ -6 } }{ x } =\dfrac { 4\times { 10 }^{ -6 } }{ x-15 }$$
$$\Rightarrow \dfrac { 6 }{ x } =\dfrac { 4 }{ x-15 }$$
$$\Rightarrow 6x-90=4x$$
$$\Rightarrow 2x=90$$
$$\Rightarrow x=\dfrac { 90 }{ 2 } =45 cm$$
A parallel plate air capacitor has capacity $$'C'$$ farad, potential $$'V'$$ volt and energy $$'E'$$ joule. When the gap between the plates is completely filled with dielectric.
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Both $$V$$ and $$E$$ increase
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Both $$V$$ and $$E$$ decrease
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$$V$$ decreases, $$E$$ increases
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$$V$$ increases, $$E$$ decreases
Explanation
A parallel- plate capacitor with a dielectric. The electric field is reduced between the plates because the dielectric material is polarized, producing an opposing field. When there is a dielectric, the potential is also reduced because potential is inversely
proportional to dielectric $$V'= \dfrac{V}{K}$$ where $$K$$ is dielectric constant.
Two place of a parallel plate capacitor of capacity $$ 50 \mu F $$ are charged by a battery to a potential of 100 V. The battery remains connected and the place are separated from each other so that the distance between them is doubled. The energy spent by the battery in doing so, will be :
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$$ 12.5 \times 10^{-2} J $$
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$$ -25 \times 10^{-2} J $$
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$$ 25 \times 10^{-2} J $$
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$$ - 12.5 \times 10^{-2} J $$
Explanation
We know that when separation between the plates is doubled the capacitance becomes one half $$ i.e., C = 25 \mu F $$
The energy spend by the battery is given by
$$ qV = (C'V)V = C'V^2 $$
$$ = 25 \times 10^{-6} \times (100)^2 $$
$$ = 25 \times 10^{-2} \ J$$
A tin nucleus has charge $$50eV$$. If the proton is $$ 10^{-12} m $$ from the nucleus. Then, the potential at this position will be :
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$$ 7.2 \times 10^8 V $$
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$$ 3.6 \times 10^4 V $$
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$$ 1.44 \times 10^4 V $$
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$$ 7.2 \times 10^4 V $$
Explanation
Charge on the nucleus
Q = 50 eV
$$ = 50 \times 1.6 \times 10^{-19} = 80 \times 10^{-19} $$
Distance of proton from nucleus
$$ r = 10^{-12} m $$
Potential of this position is given by
$$ V = \dfrac {1}{4 \pi l_0 } . \dfrac {Q}{r} $$
$$ = 9 \times 10^9 \times \dfrac {80 \times 10^{-19}}{10^{-12}} $$
$$ = 7.2 \times 10^4 V $$
A $$2 \mu F$$ capacitor is charged to $$100 V$$ and then its plates are connected by a conducting wire, the heat produced is :
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$$1 J$$
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$$0.1 J$$
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$$0.01 J$$
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$$0.001 J$$
Explanation
Here $$C = 2 \mu F$$, $$V = 100$$ volt
Heat produced $$H = \dfrac { 1 }{ 2 } C{ V }^{ 2 }$$
$$= \dfrac { 1 }{ 2 } \left( 2\times { 10 }^{ -6 } \right) \times { \left( 100 \right) }^{ 2 }$$
$$= 1 \times { 10 }^{ -2 } J$$
$$= 0.01 J$$
A $$16pF$$ capacitor is connected to $$70V$$ supply. The amount of electric energy stored in the capacitor is:
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$$4.5\times { 10 }^{ -12 }J\quad $$
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$$5.1\times { 10 }^{ -8 }J$$
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$$2.5\times { 10 }^{ -12 }J$$
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$$3.2\times { 10 }^{ -8 }J$$
Explanation
Here, $$C=16\mu F=16\times { 10 }^{ -12 }F;V=80V$$
As $$V=\cfrac { 1 }{ 2 } C{ V }^{ 2 }=\cfrac { 1 }{ 2 } \times 16\times { 10 }^{ -12 }\times { (80) }^{ 2 }=5.1\times { 10 }^{ -8 }J$$
Three charges $$ 1 \mu C, 2 \mu C, $$ and $$ 3 \mu C $$ are kept at vertices of an equilateral triangle of side 1 m. If they are brought nearer, so that they now form an equilateral triangle of side $$0.5\, m$$, then work done is :
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$$0.11 \,J$$
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$$11 \,J$$
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$$0.01 \,J$$
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$$1.1 \,J$$
Explanation
Initial PE of the three charges,
$$ U_1 = k \dfrac {q_1 q_2 + q_2 q_3 + q_1 q_3 }{r} $$
$$ =9\times 10^{ 9 }\begin{bmatrix} \dfrac { 1\times 2\times 10^{ -12 }+2\times 3\times 10^{ -12 }+ 1 \times 3 \times 10^{ -12 } }{ 1 } \end{bmatrix}$$
$$ = 99 \times 10^{-3} J $$
Final PE of the system
$$ U_f = 9 \times 10^9 \begin{bmatrix} \dfrac { 1\times 2\times 10^{ -12 }+2\times 3\times 10^{ -12 }+ 1 \times 3 \times 10^{ -12 } }{ 0.5 } \end{bmatrix}$$
$$ = \dfrac {99 \times 10^{-3} } { 0. 5 } = 198 \times 10^{-3} J $$
$$ W = U_f - U_i = ( 198 - 99 ) \times 10^{-3} J $$
$$ = 99 \times 10^{-3} J $$
$$ = 0.099 = 0.01 J $$
A capacitor of $$ 10 \mu F $$ charged upto 250 V is connected in parallel with another capacitor of $$ 5 \mu F $$ charged upto 100 V. The common potential is :
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200 V
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300 V
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400 V
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500 V
Explanation
The equivalent capacitance of the capacitors joined in parallel is equal to the sum of their individual capacitances.
Let the charge on the capacitors be $$q_1$$ and $$q_2 . $$ Then total charge
$$ Q = Q + Q_2 $$
$$ CV = C_1 V_1 + C_2 V_2 $$
Since, capacitors are connected in parallel equivalent capacitance is $$ C = C_1 + C_2 $$
$$ V = \dfrac {C_1 V_1 + C_2 V_2}{C_1 + C_2 } $$
Given, $$ C_1 = 10 \mu F, V_1 - 250 V, C_2 = 5 \mu F , V_2 = 100 V $$
$$ \therefore V = \dfrac {(10 \times 10^{-6} \times 250 )+(5 \times 10^{-6} \times 100)}{(10 \times 10^{-6} + 5 \times 10^{-6})} $$
$$ \Rightarrow V = \dfrac {3000 \times 10^{-6}}{15 \times 10^{-6}} = 200 V $$
The potential of a large liquid drop when eight liquid drops are combined is $$20\ V$$. Then, the potential of each single drop was :
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$$10\ V$$
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$$7.5\ V$$
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$$5\ V$$
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$$2.5\ V$$
Explanation
Volume of $$8$$ drops $$=$$ Volume of big drop
$$\therefore \left (\dfrac {4}{3}\pi r^{3}\right ) \times 8 = \dfrac {4}{3} \pi R^{3}$$
$$\Rightarrow 2r = R .... (i)$$
According to charge conservation
$$8q = Q .... (ii)$$
We have
$$V \propto \dfrac {1}{r}$$
or $$\dfrac {V_{1}}{V_{2}} = \dfrac {r_{2}}{n}$$
$$\Rightarrow \dfrac {20}{V_{2}} = \dfrac {2r}{r}$$
$$\Rightarrow V_{2} = 10\ V$$
Capacitance of a capacitor becomes $$ \frac {7}{6} $$ times its original value, if a dielectric slab of thickness $$ t = \frac {2}{3} d $$ is introduced in between the plates. The dielectic constant of the dielectric slab is :
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$$\frac{14}{11} $$
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$$\frac{11}{14} $$
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$$\frac{7}{11} $$
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$$\frac{11}{7} $$
Explanation
The capacitance is given by $$ C_1 = \frac {\epsilon_0 A }{d} $$ ...(i)
If electric slab of constant k of thickness $$ t = \frac {2}{3} d $$ is introduced, then capacitance becomes
$$ C_2 = \frac {\epsilon_0 A}{d - t \left( 1 - \frac {1}{k} \right) } $$
$$ = \frac {\epsilon_0 A }{d - \frac{2}{3} d \left( 1 - \frac {1}{k} \right) } $$ ...(ii)
Eq. (i) becomes
$$ C_2 = \frac {\epsilon_0 A }{d \begin{bmatrix} \left( 1 - \frac {2}{3} \right) + \frac {2}{3k} \end{bmatrix} } $$
$$\Rightarrow C_{ 2 }=\frac { C_{ 1 } }{ \left( \frac { 1 }{ 3 } +\frac { 2 }{ 3k } \right) } $$
$$ \Rightarrow \frac {7}{6} C_1 = \frac {C_1}{\frac {1}{3} + \frac {2}{3k}} $$
$$ \Rightarrow \frac {1}{3} + \frac {2}{3k} = \frac {6}{7} $$
$$ \Rightarrow \frac {2}{3k} = \frac {6}{7} - \frac {1}{3} = \frac {11}{21} $$
or $$ 33k = 21 \times 2 $$
$$ k = \frac {42}{33} = \frac {14}{11} $$
$$A \ \ 6 \times 10^{-4}$$ F parallel plate air capacitor is connected to a 500 V battery. When air is replaced by another dielectric material, $$7.5 \times 10^{-4} C$$ charge flows into the capacitor. The value of the dielectric constant of the material is:
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1.5
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2.0
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1.0025
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3.5
Explanation
Let the dielectric constant be $$K$$.
Capacitance of parallel plate air capacitor $$C = 6\times 10^{-4} \ F$$
Voltage of the battery $$V =500$$ volts
Thus charge stored on air capacitor $$Q_a = CV = (7.5\times 10^{-4})(500) = 3000 \times 10^{-4} $$ C
Extra charge flown $$Q = 7.5\times 10^{-4} $$ C
Thus total charge stored on dielectric capacitor $$Q' = (3000+7.5)\times 10^{-4} = 3007.5\times 10^{-4}$$ C
Thus capacitance of parallel plate dielectric capacitor $$C = \dfrac{Q'}{V} = \dfrac{3007.5\times 10^{-4}}{500} =6.015\times 10^{-4} \ F$$
Capacitance of parallel plate dielectric capacitor $$C' = KC$$
$$\therefore$$ $$.015\times 10^{-4} = K(6\times 10^{-4})$$
$$\implies \ K = 1.0025$$
A conductor with a cavity is charged positively and its surface charge density is $$\sigma$$. If E and V represent the electric field and potential, then inside the cavity:
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and $$V=0$$
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$$E=0$$ and $$V=0$$
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$$E=0$$ and $$\sigma =$$constant
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$$V=0$$ and $$\sigma =$$ constant
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$$E=0$$ and $$V=$$ constant
Explanation
Surface charge density $$=\sigma$$.
Inside the surface, the electric field is zero.
$$\Rightarrow E_{inside}=0$$
$$\because V_{inside}=V_{surface}$$
The potential inside the surface of charged sphere is constant.
$$\therefore E=0, V=$$constant.
Choose the wrong statement about equipotential surfaces.
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It is a surface over which the potential is constant
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The electric field is parallel to the equipotential surface
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The electric field is perpendicular to the equipotential surface
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The electric field is in the direction of steepest decrease of potential
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They are concentric spheres for a point charge
Explanation
Equipotential lines are curved lines on a map which mark out lines of identicalaltitude. The altitude pertains to electric potential or voltage. Equipotential lines are always perpendicular to the electric field. The lines creates equipotential surfaces in a three dimensions. Movement along an equipotential surface needs no work since such movement is always perpendicular to the electric field. The figure below shows the equipotential surfaces in dashed lines and electric field lines in solid lines produced by a positive point charge. In this case, the equipotential surfaces are spheres are on the center of the charge.
Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch S. If now the switch S is opened and the space between the plates is filled with a dielectric of relative permittivity $$\displaystyle { \varepsilon }_{ r }$$, then :
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The potential difference as well as charge on each capacitor goes up by a factor $$\displaystyle { \varepsilon }_{ r }$$
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The potential difference as well as charge on each capacitor goes down by a factor $$\displaystyle { \varepsilon }_{ r }$$
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The potential difference across A remains constant and the charge on B remains unchanged
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The potential difference across B remains constant while the charge on A remains unchanged
Explanation
After switch S is opened, as the Capacitor A is connected across the battery, its potential difference is fixed at steady state (i.e., when capacitor is fully charged).
Capacitor B is isolated, so its charge gets fixed. But as we insert the dielectric, its capacitance changes, thus its potential difference also changes.
Option C is correct.
Four capacitors are joined as shown in the adjoining figure. The capacitance of each is $$ 8\mu F $$. The equivalent capacitance between the points A and B is
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$$ 32\mu F $$
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$$ 2\mu F$$
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$$ 8\mu F $$
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$$ 16\mu F $$
Three capacitors connected in series have an effective capacitance of $$4 \mu F$$. If one of the capacitance is removed, the net capacitance of the capacitor increases to $$6 \mu F$$. The removed capacitor has a capacitance of
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$$2\mu F$$
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$$4\mu F$$
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$$10\mu F$$
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$$12\mu F$$
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$$24\mu F$$
Explanation
Let there are three capacitors with capacitances $$C_1,C_2 , C_3$$ respectively and $$ C_1 $$ is removed.
In first case, $$\dfrac{1}{C_{eq1}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3} $$ ...(1)
In second case,
$$\dfrac{1}{C_{eq2}}=\dfrac{1}{C_2}+\dfrac{1}{C_3} $$ ...(2)
From (1) and (2),
$$\dfrac{1}{C_{eq1}}=\dfrac{1}{C_1}+\dfrac{1}{C_{eq2}} $$
or $$1/4=1/C_1+1/6$$
or $$C_1=12\, \mu F$$
Capacitance of a capacitor made by a thin metal foil is $$\displaystyle 2\mu F$$. If the foil is folded with paper of thickness 0.15 mm, dielectric constant of paper is 2.5 and width of paper is 400 mm, the length of the foil will be :
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0.34 m
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1.33 m
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13.4 m
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33.9 m
Explanation
Here, let the length be $$l$$
width is, $$b=400mm=0.4m$$
thickness, $$t=0.15mm=15\times10^{-5}m$$
As capacitance for parallel plate capacitor is,
$$C=K\epsilon_0A/d=K\epsilon_0(l\times b)/t$$
$$\implies l=Ct/K\epsilon_0b=2\times10^{-6}\times15\times10^{-5}/(2.5\times8.85\times10^{-12}\times0.4)=33.91m$$
Option D is correct.
If a dielectric introduced between the plates of a parallel plate condenser, then which of the following is possible :
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Decreases the electric field between the plates
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Decreases the capacity of the condenser
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Increases the charge stored in the condenser
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Increases the capacity of the condenser
Explanation
If a dielectric medium of dielectric constant $$K$$ is filled completely between the plates then capacitance increases by $$K$$ times
$$C' = \dfrac {K\epsilon_{0}A}{d}$$
$$\Rightarrow C' = KC$$.
An uncharged parallel plate capacitor filled with a dielectric of dielectric constant K is connected to an air filled identical parallel capacitor charged to potential $$V_1$$. If the common potential is $$V_2$$, the value of K is
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$$\dfrac{V_1-V_2}{V_1}$$
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$$\dfrac{V_1}{V_1-V_2}$$
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$$\dfrac{V_2}{V_1-V_2}$$
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$$\dfrac{V_1-V_2}{V_2}$$
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$$\dfrac{V_1-V_2}{V_1+V_2}$$
Explanation
If the capacitance of the air filled capacitor is $$C_1=C$$, then the capacitance of the dielectric field capacitor will be $$C_2=KC$$
Now, the common potential is $$V_2=\dfrac{Q_1+Q_2}{C_1+C_2}$$
Here, $$Q_1=C_1V_1$$, charge for air filled capacitor and as the dielectric filled capacitor is uncharged, $$Q_2=0$$
Thus, $$V_2=\dfrac{C_1V_1+0}{C_1+C_2}$$
or $$V_2=\dfrac{CV_1}{C+KC}$$
or $$K+1=V_1/V_2$$
or $$K=\dfrac{V_1-V_2}{V_2}$$
The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is $$\displaystyle 2\mu F$$. The separation is reduced to half and it is filled with a dielectric substance of value 2.The final capacity of the capacitor is
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$$\displaystyle 11.2\mu F$$
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$$\displaystyle 15.6\mu F$$
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$$\displaystyle 19.2\mu F$$
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$$\displaystyle 22.4\mu F$$
Explanation
Initial capacitance $$\displaystyle C = 2uF=\frac { { \varepsilon }_{ 0 }{ A }}{ d } $$
New capacitance $$\displaystyle { C }^{ \prime }=\frac { K{ \varepsilon }_{ 0 }A }{ { d }^{ \prime } } =\frac { 2.8\times { \varepsilon }_{ 0 }A }{ (d/2) } =5.6\frac { { \varepsilon }_{ 0 }A }{ d } $$
$$\displaystyle \Rightarrow { C }^{ \prime }= 5.6\times 2=11.2\mu F$$
A capacitor of capacitance $$10\mu F$$ is charged by connecting through a resistance of $$20\Omega$$ and a battery of $$20V$$. What is the energy supplied by the battery
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Less than $$2 m \,J$$
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$$2 m \,J$$
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More than $$2 m \,J$$
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Cannot be predicted
Explanation
Given,
$$ C = 10\ \mu F $$
$$ R = 20\ \Omega $$
$$ V = 20\ V $$
We know, current in an RC circuit is given by
$$ i = \dfrac{V}{R} e^\left({\dfrac{-t}{RC}} \right) $$
Power dissipated by the resistor is given by
$$ P_R = i^2R = \left( \dfrac{V}{R} e^\left( {\dfrac{-t}{RC}}\right) \right)^2 R = \dfrac{V^2}{R} e^\left({\dfrac{-2t}{RC}} \right) $$
Energy consumed by the resistor $$ E_R = \int_0^\infty P_R dt$$
$$ E_R = \displaystyle \int_0^\infty \dfrac{V^2}{R} e^\left({\dfrac{-2t}{RC}} \right)\ dt $$
$$ E_R = \dfrac{V^2}{R} \displaystyle \int_0^\infty \ e^\left({\dfrac{-2t}{RC}} \right)\ dt $$
$$ E_R = \dfrac{V^2RC}{R2} \displaystyle | e^0 - e^\infty | = \dfrac{1}{2} CV^2 $$
$$ E_R = \dfrac{1}{2} \times 10 \times 10^{-6} \times 20^2 = 2 \times 10^{-3} J = 2\ mJ $$
Energy consumed by resistor alone is equal to $$ 2\ mJ $$
$$ \therefore$$ Energy supplied by battery must be $$ \gt 2\ mJ $$
Hence, the correct answer is OPTION C.
The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $$0.4cm$$ is $$2\mu F$$. If the separation is reduced to half and it is filled with a dielectric substance of value $$2.8$$, then the final capacity of the capacitor is
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$$11.2\mu F$$
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$$15.6\mu F$$
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$$19.2\mu F$$
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$$22.4\mu F$$
Explanation
Given, $$C=\cfrac { { \varepsilon }_{ 0 }A }{ d } =2\mu F$$
and $$C'=\cfrac { K{ \varepsilon }_{ 0 }A }{ d' } =\cfrac { 2.8{ \varepsilon }_{ 0 }A }{ d/2 } =\cfrac { 5.6{ \varepsilon }_{ 0 }A }{ d } $$
$$\Rightarrow C'=5.60\times 2=11.2\mu F$$
A capacitor stores $$40$$ $$\mu$$C charge when connected across a battery. When the gap between the plates is filled with a dielectric a charge of $$80$$ $$\mu$$C flows through the battery. The dielectric constant of dielectric inserted is?
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$$2$$
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$$3$$
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$$1$$
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$$4$$
Explanation
$$\because \displaystyle C_0=\frac{\varepsilon_0A}{d}$$ and $$C=\displaystyle\frac{\varepsilon A}{d}=\frac{K\varepsilon_0A}{d}$$
So, $$\displaystyle q_0=C_0V=\frac{\varepsilon A}{d}V$$
and $$\displaystyle q=CV=\frac{K\varepsilon_0A}{d}V$$
$$\therefore \displaystyle\frac{q}{q_0}=K$$
Given $$q_0=40\mu C$$
On inserting a dielectric, an additional charge of $$80\mu C$$ also flows through the battery.
So, total charge $$=q_0+80=40+80=120\mu C$$
$$\therefore K=\displaystyle\frac{q}{q_0}=\frac{120}{40}=3$$.
Two capacitors of capacitance $$C$$ are connected in series. If one of them is filled with a substance of dielectric constant $$K$$, what is the effective capacitance?
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$$\cfrac { KC }{ \left( 1+k \right) } $$
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$$C(K+1)$$
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$$\cfrac { 2KC }{ \left( 1+k \right) } $$
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None of these
Explanation
When a substance of dielectric constant $$K$$ is filled in the capacitor, its capacity
$${ C }_{ m }=\cfrac { K{ \varepsilon }_{ 0 }A }{ d } =K{ C }_{ o }$$
Here, $${ C }_{ o }=C$$
$$\therefore { C }_{ m }==KC$$
Now two capacitors of capacities $$KC$$ and $$C$$ are connected in series, therefore their effective capacitance
$$\cfrac { 1 }{ C' } =\cfrac { 1 }{ KC } +\cfrac { 1 }{ C } =\cfrac { 1+K }{ KC } \Rightarrow C'=\cfrac { KC }{ (1+K) } $$
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