MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 4

Figure shows three points A, B and C in a region of uniform electric field

$\vec E$ . The line AB is perpendicular and BC is parallel to the field lines. Then which of the following holds good. Here

$V_A, V_B$ and

$V_C$ represents the electric potential at point A, B and C respectively.

0%
$V_A=V_B=V_C$
0%
$V_A=V_B > V_C$
0%
$V_A=V_B < V_C$
0%
$V_A > V_B = V_C$

Explanation

As we know the electrical lines of force always travel from higher potential to lower potential and lines of forces are perpendicular to the equipotential surface. So,

$V_A=V_B$ and

$V_B>V_C$

Find the charge appearing on capacitor of capacitance

$4\;\mu F$ :

0%
$32\;\mu C$
0%
$96\;\mu C$
0%
$12\;\mu C$
0%
$4\;\mu C$

Explanation

As the two capacitances are in series combination, so equivalent capacitance,

$C_{equ}=\dfrac{8\times4}{8+4}=\dfrac{32}{12}=\dfrac{8}{3} \mu F$
So, charge on

$4\ \mu F, Q=C_{equ}\times V=\dfrac{8}{3}\times12=32 \ \mu C$

A capacitor stores

$60\space\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of

$120\space\mu C$ flows through the battery. The dielectric constant of the material inserted is :

0%
$1$
0%
$2$
0%
$3$
0%
none

Explanation

The capacitance before filled with dielectric is

$C=\dfrac{A\epsilon_0}{d}$
The capacitance after filled with dielectric is

$C'=\dfrac{Ak\epsilon_0}{d}=kC$
Initial charge,

$Q=60 \mu C$ and total charge after filled with dielectric

$Q'=120+60=180 \mu C$
Here,

$Q=CV$ and

$Q'=C'V=kCV$

$\dfrac{Q'}{Q}=\dfrac{kCV}{CV}$

$k=\dfrac{Q'}{Q}=\dfrac{180}{60}=3$

In the adjoining figure, capacitor

$(1)$ and

$(2)$ have a capacitance

$'C'$ each. When the dielectric of dielectric constant

$K$ is inserted between the plates of one of the capacitor, the total charge flowing through battery is :

0%
$\displaystyle\frac{KCE}{K+1}\space from\space B\space to\space C$
0%
$\displaystyle\frac{KCE}{K+1}\space from\space C\space to\space B$
0%
$\displaystyle\frac{(K-1)CE}{2(K+1)}\space from\space B\space to\space C$
0%
$\displaystyle\frac{(K-1)CE}{2(K+1)}\space from\space C\space to\space B$

Explanation

Before insert dielectric, the equivalent capacitance

$C_{eq}= \frac{C\times C}{C+C}=C/2$

$Q_{eq}=C_{eq}E=CE/2$
As battery is connected to capacitor so E remains constant.
After insert dielectric ,

$C'_{eq}=\dfrac{KC C}{KC+C}=\dfrac{K}{K+1}C$

$Q'_{eq}=C'_{eq}E=\dfrac{K}{K+1}CE$
Charge flow through the battery is

$\Delta Q=Q'_{eq}-Q_{eq}=\frac{K}{K+1}CE-\dfrac{CE}{2}=\dfrac{(K-1)CE}{2(K+1)}$
As E is constant so for inserting dielectric the charge on the capacitor will increase and it will flow from capacitance to battery to maintain constant

$E$

Four identical plates

$1, 2, 3,$ and

$4$ are placed parallel to each other at equal distance as shown in the figure. Plates

$1$ and

$4$ are joined together and the space between

$2$ and

$3$ is filled with a dielectric of dielectric constant

$k = 2$ . The capacitance of the system between

$1$ and

$3$ &

$2$ and

$4$ are

$C_1$ and

$C_2$ respectively. The ratio

$C_1/C_2$ is :

0%
$\displaystyle\frac{5}{3}$
0%
$1$
0%
$\displaystyle\frac{3}{5}$
0%
$\displaystyle\frac{5}{7}$

Explanation

The capacitance of parallel plate capacitor is

$C=\dfrac{A\epsilon_0}{d}$
The capacitance of parallel plate capacitor with dielectric is

$C'=\dfrac{Ak\epsilon_0}{d}=kC=2C$
Here 1 and 2 plate and 3 and 4 plate will make the capacitor of same capacitance

$C$ and 2 and 3 will make the capacitance of

$C'$ .
Here the capacitance in between 1 and 3 is

$C_1=\dfrac{CC'}{C+C'}=\dfrac{C.2C}{C+2C}=2C/3$
and the capacitance in between 2 and 4 is

$C_2=\dfrac{CC'}{C+C'}=\dfrac{C.2C}{C+2C}=2C/3$

$\therefore \frac{C_1}{C_2}=1$

A capacitor of capacitance

$C$ is initially charged to a potential difference of

$V$ volt. Now it is connected to a battery of

$2V$ volt with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be :

0%
$1.75$
0%
$2.25$
0%
$2.5$
0%
$\dfrac{1}{2}$

Explanation

Here

$q_i=CV$ and

$q_f=2CV$
As battery is connected with opposite polarity so charge flow through the battery is

$\Delta q=q_f-(-q_i)=2CV+CV=3CV$ .
Initial energy stored ,

$U_i=\dfrac{1}{2}CV^2$ and
Final energy stored ,

$U_f=\dfrac{1}{2}C(2V)^2=2CV^2$ .
Generated heat,

$H=(U_f-U_i)-\Delta q(2V)=2CV^2-\dfrac{1}{2}CV^2-6CV^2=-\dfrac{9}{2}CV^2$
Thus,

$\dfrac{H}{U_f}=\dfrac{9CV^2}{2(2CV^2)}=2.25$

The capacitance of a parallel plate capacitor is

$C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant

$k$ . The capacitor is connected to a cell of emf

$E$ , and the slab is taken out

0%
charge $CE(k-1)$ flows through the cell
0%
energy $E^2C(k-1)$ is absorbed by the cell
0%
the energy stored in the capacitor is reduces by $E^2C(k-1)$
0%
the external agent has to do $\displaystyle\frac{1}{2}E^2C(k-1)$ amount of work to take the slab out

Explanation

After inserting the dielectric the capacitance will be

$C'=kC$
The charge flow through the cell when the slab insert is

$Q'=C'E=kCE$
The charge flow through the cell when the slab is taken out

$=Q=CE$
Net charge flow through the cell

$Q_n=Q'-Q=kCE-CE=CE(k-1)$
The amount of work for taking out slab is

$W=\dfrac{1}{2}(CkE^2-CE^2)=\dfrac{1}{2}CE^2(k-1)$

Two capacitors

$C_1$ and

$C_2$ are connected in series, assume that

$C_1 < C_2$ . The equivalent capacitance of this arrangement is

$C$ , where

0%
$C < C_1/2$
0%
$C_1/2 < C < C_1$
0%
$C_1 < C < C_2$
0%
$C_2 < C < 2C_2$

Explanation

We have series capacitance,

$\frac {1}{C} = \frac {1}{C_1} + \frac {1}{C_2}$

$\implies C = \frac {C_1 C_2}{C_1 + C_2}$

And also given that

$C_2 > C_1$

Hence,

$2C_2 > C_2 > C_1 > C > \frac{C_1}{2}$

Option

$(a)$ ,

$(c)$ and

$(d)$ are wrong.

Statement 1: The electrostatic force between the plates of a charged isolated capacitor decreases when dielectric fills whole space between plates.

Statement 2: The electric field between the plates of a charged isolated capacitance decreases when dielectric fills whole space between plates

0%
Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1
0%
Statement-1 is true, Statement-2 is true and Statement-2 is NOT the correct explanation for Statement-1
0%
Statement-1 is true, Statement-2 is false
0%
Statement-1 is false, Statement-2 is true

Explanation

For isolated capacitor the charge (Q) remains constant.
The field between the plates without dielectric is

$E=\frac{Q}{A\epsilon_0}$
The field between the plates with dielectric is

$E'=\frac{Q}{Ak\epsilon_0}=E/k$ where k is dielectric constant. Thus the filed between the plates will decrease.
The force between the plates do not depend on the field between the plates and its depend on the filed due to a charge plate.

A capacitor of capacitance

$1 \mu F$ withstands a maximum voltage of 6 kilovolt while another capacitor of

$2 \mu F$ withstands a maximum voltage 4 kilovolt . if the two capacitor are connected in series, the system will withstand a maximum of:

0%
2kV
0%
4kV
0%
6kV
0%
9kV

A parallel plate capacitor has a parallel slab of copper inserted between and parallel to the two plates, without touching the plates. The capacity of the capacitor after the introduction of the copper sheet is:

0%
minimum when the copper slab touches one of the plates
0%
maximum when the copper slab touches one of the plates
0%
invariant for all positions of the slab between the plates
0%
greater than that before introducing the slab

Explanation

As the copper is a good conductor, the charge on the capacitor will appear on the large flat surfaces of the copper sheet, with the negative side of the copper facing the positive side of the capacitor. This arrangement can be considered to be two capacitors in series.
If thickness of the copper slab is

$l$ , the separation of plates between will be

$\dfrac{1}{2}(d-l)$
The capacitance before insert slab is

$C=\dfrac{A\epsilon_0}{d}$
The capacitance of each capacitor after insert slab is

$C'=\dfrac{A\epsilon_0}{(d-l)/2}$
Net capacitance after insert slab is

$C_{eq}=\dfrac{C'C'}{C'+C'}=\frac{C'}{2}=\dfrac{A\epsilon_0}{(d-l)}$
thus,

$C_{eq}>C$ and if sheet is very thin

$(d>>l), C_{eq}=C$

A dense sphere of mass

$M$ is placed at the centre of a circle of radius

$R$ . Find the work done, when a particle of mass

$m$ is brought from A to B along a circle as shown in the figure.

0%
$zero$
0%
$\displaystyle \frac{G M m}{R}$
0%
$\displaystyle - \frac{G Mm}{R}$
0%
$\displaystyle \frac{2 G M m}{R}$

Explanation

$E = PE + KE$
From energy conservation,

$\dfrac{1}{2}mv_A ^2 + (PE)_A= \dfrac{1}{2}mv_B ^2 + (PE)_B$
Since speed does not change from A to B, KE remains same.
Distance of the particle at A and B is equal to the radius of the the circle.

$PE = -\dfrac{GMm}{R}$
Hence,

$(PE)_A = (PE)_B$
Hence, work done is zero as there is no change in total energy E from A to B.

A capacitor of capacitance 10

$\mu$ F is charged by connecting through a resistance of 20 ohm and a battery of 20 V. What is the energy supplied by the battery?

0%
Less than 2 m J
0%
2 m J
0%
More than 2 m J
0%
Cannot be predicted

Explanation

The energy supplied by the battery will be

$P=\dfrac{V^2}{R}=\dfrac{20^2}{20}=20 J$
So it is more than

$2 mJ$

Potential in electrostatics is analogous to:

0%
pressure in gases
0%
temperature in thermal physics
0%
levels in liquids
0%
all of the above

P is a point on an equipotential surface S. The field at P is E.

0%
E must be perpendicular to S in all cases.
0%
E will be perpendicular to S only if S is a plane surface.
0%
E cannot have a component along a tangent to S.
0%
E may have a nonzero component along a tangent to S if S is a curved surface.

An electrical charge of

$2$

$\mu$ C is placed at the point

$(1, 2, 3)$ . At the point

$(2, 3, 4)$ the electric field and potential will be :

0%
$6 \times 10^3 NC^{-1}$ and $6 \times 10^3 JC^{-1}$
0%
$6000 NC^{-1}$ and $6000 \sqrt{3} JC^{-1}$
0%
$6 \times 10^3 NC^{-1}$ and $3\sqrt{3} JC^{-1}$
0%
none of the above

Explanation

The electric field , $\displaystyle E=\dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r^2}=9\times 10^9\times \dfrac{2\times 10^{-6}}{(2-1)^2+(3-2)^2+(4-3)^2}=6000 NC^{-1}$

and potential, $\displaystyle V=\dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r}=9\times 10^9\times \dfrac{2\times 10^{-6}}{\sqrt{(2-1)^2+(3-2)^2+(4-3)^2}}=6000 \sqrt 3 NC^{-1}$

An air-filled parallel plate capacitor has a capacity

$2 pF$ . The separation between the plates is doubled and the interspace is filled with wax, If the capacity is increased to 6 pF, the dielectric constant of the wax is :

0%
$2$
0%
$4$
0%
$3$
0%
$6$

Explanation

when Air filled , capacitance

$=C=2 pF$
As C is inversely proportional to the separation so it is double the capacitance will be

$C/2=$
If dielectric constant is

$k$ , after filling wax the capacitance becomes

$C'=k\dfrac{C}{2}$
now

$k=\dfrac{C'}{C/2}=\dfrac{6}{2/2}=6$

In a uniform electric field, equipotential surfaces must :

0%
be plane surfaces
0%
be normal to the direction of the field
0%
be spaced such that surfaces having equal differences in potential are separated by equal distances
0%
have decreasing potentials in the direction of the field

Explanation

Uniform field means a constant field at every point and it is directed in a fixed direction for all points. Thus equipotential surfaces must be plane surface. The electric field is always perpendicular to equipotential surface because

$\vec{E}=-\vec{\nabla}{V}$ . As surfaces are equipotential so it must be spaced such that surfaces having equal differences in potential are separated by equal distances. As

$E=-\dfrac{dV}{dr}$ so it must have decreasing potentials in the direction of the field.

Two equal charges A and B each of

$1/3 \times 10^{-6}C$ are placed 200 cm apart in air. A particle carrying a charge of

$-1/3 \times 10^{-6}C$ is projected along the perpendicular bisector from the point O midway between A and B with a kinetic energy of

$10^{-3} J$ . Before the particle starts to return it will cover a distance

0%
$1 m$
0%
$\sqrt{2}$ m
0%
$\sqrt{3}$ m
0%
$1/\sqrt{3}$ m

Explanation

Initial potential energy: $\dfrac { 2k{ q }_{ 1 }{ q }_{ 2 } }{ r } =\quad (2\times 9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 3 } \times { 10 }^{ -6 }\times \dfrac { -1 }{ 3 } \times { 10 }^{ -6 })/(1)=-2\times { 10 }^{ -3 }$

Initial kinetic energy is 10^(-3) J.

Let the body is at r distance from both the charges (will be at equal distance from both as both charges are same)

Final Kinetic energy is zero (as body stopped).

By equating Initial Total Energy & Final Total Energy:

$-2\times { 10 }^{ -3 }+{ 10 }^{ -3 }=\quad \dfrac { 2k{ q }_{ 1 }{ q }_{ 2 } }{ r } =(2\times 9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 3 } \times { 10 }^{ -6 }\times \dfrac { -1 }{ 3 } \times { 10 }^{ -6 })/(r)\\ -1\times { 10 }^{ -3 }=-2\times { 10 }^{ -3 }/(r)\\ \therefore \quad r=2$

Therefore distance traveled:

$\sqrt { { ({ r }_{ f }^{ 2 }- }{ r }_{ i }^{ 2 }) } =\sqrt { 4-1 } =\sqrt { 3 }$

A capacitor has charge

$50\mu$ C. When the gap between the plates is filled with glass wool 120

$\mu$ C charge flows through the battery. The dielectric constant of glass wool is :

0%
$3.4$
0%
$1.4$
0%
$2.4$
0%
none of these

Explanation

Before filled wool the charge on capacitor is

$Q$
After filled wool the charge on capacitor is

$Q'=kQ$ where

$k=$ dielectric constant.
So charge flow through the battery is

$Q'-Q=120 \Rightarrow Q'=120+Q=120+50=170$
thus,

$kQ=170 \Rightarrow k=\dfrac{170}{50}=3.4$

The energy stored in a condenser is in the form of

0%
electrostatic energy
0%
magnetic energy
0%
elastic energy
0%
kinetic energy

Explanation

Energy in a condenser is always stored in the form of electrostatic energy (

$U$ ) and it is given by

$U = \frac {1}{2} \frac{Q^2}{C}$
where

$Q$ is the total charge and

$C$ is the capacitance of the condenser.

Potential of point A is?

0%
$3$ V
0%
$6$ V
0%
$9$ V
0%
Zero

The capacitance of a parallel-plate capacitor is

$C_0$ when the region between the plates has air. This region is now filled with a dielectric slab of dielectric constant K. The capacitor is connected to a cell emf

$\varepsilon _1$ and the slab is taken out.

0%
Charge $\varepsilon C_0(K-1)$ flows through the cell.
0%
Energy $\varepsilon ^2C_0(K-1)$ is absorbed by the cell.
0%
The energy stored in the capacitor is reduced by $\varepsilon ^2C_0(K-1)$ .
0%
The external agent has to do $\frac{1}{2}\varepsilon ^2C_0(K-1)$ amount of work to take the slab out.

Explanation

The capacitance of capacitor with dielectric is

$C=KC_0$
Initial charge on capacitor is

$q=C\varepsilon=KC_0\varepsilon$
As cell is still connect so potential remain constant.
after slab is taken out the charge on capacitor is

$q'=C_0\varepsilon$
The charge flows through the cell is

$\Delta q=q-q'=KC_0\varepsilon-C_0\varepsilon=C_0\varepsilon(K-1)$
Energy absorbed by cell

$E=\Delta q \varepsilon=C_0\varepsilon^2(K-1)$
Energy stored in capacitor

$U=\frac{1}{2}q\varepsilon-\frac{1}{2}q'\varepsilon=\frac{1}{2}C_0\varepsilon^2(K-1)$
Work done

$W=E-U=C_0\varepsilon^2(K-1)-\frac{1}{2}C_0\varepsilon^2(K-1)=\frac{1}{2}C_0\varepsilon^2(K-1)$

$1\ mm$ thick paper of dielectric constant 4 lies between the plates of a parallel-plate capacitor. It is charged to 100 volt the intensity of electric field between the plates of the condenser will be :

0%
100
0%
100000
0%
400000
0%
25000

Explanation

The electric field in between the plates in the presence of dielectric is

$E=\displaystyle\frac{V}{kd}=\displaystyle\frac{100}{4\times 10^{-3}}=25000$ V/m

A parallel plate condenser is connected to a battery of emf 4 volt. If a plate of dielectric constant 8 is inserted into it, the potential difference on the condenser will be :

0%
32 V
0%
4 V
0%
1/2 V
0%
2 V

Explanation

Total charged is conserved in the condenser. Here the potential difference on the condenser is equal to the emf of battery. When dielectric is inserted between the plates the charge will maintain the constant potential in the capacitor. Thus, the potential difference on the condenser is

$4 V$

The distance between the plates of a parallel plate air condenser is d. if a copper plate of the same area but thickness

$\dfrac d2$ is placed between the plates then the new capacitance will become :

0%
doubled
0%
half
0%
one fourth
0%
remain unchanged

Explanation

Capacitance of parallel plate capacitor is

$C=\dfrac{A\epsilon}{d}$
If distance between plate is

$d/2$
so new capacitance is

$C'=\dfrac{A\epsilon}{d/2}=2\dfrac{A\epsilon}{d}=2C$

A parallel-plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by

$Q_0, V_0, E_0$ and

$U_0$ respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related with previous ones as :

0%
$V > V_0$
0%
$U > U_0$
0%
$Q > Q_0$
0%
$E > E_0$

Explanation

As the capacitor is still connect with battery so the potential will remain constant as well as field and the capacitor will be charging up. thus,

$V=V_0 , E=E_0, Q>Q_0$
As the dielectric insert between the plate so the energy

$U_0=\displaystyle\frac{1}{2} CV_0^2$ becomes

$U=\displaystyle\frac{1}{2} kCV_0^2$ . thus energy will increase.

$U>U_0$

On increasing the plate separation of charged condenser its energy :

0%
remains unchanged
0%
decreases
0%
increases
0%
none of these

Explanation

Parallel plate capacitance ,

$C=\dfrac{A\epsilon_0}{d}$
Potential between the plate is

$V=\dfrac{Qd}{A\epsilon_0}$
Energy

$U=\dfrac{1}{2}CV^2=\dfrac{1}{2}\times \dfrac{A\epsilon_0}{d}\times (\dfrac{Qd}{A\epsilon_0})^2$
thus,

$U \propto d$
So when the separation d increases, energy U will increase.

On removing the dielectric from a charged condenser, its energy

0%
increase
0%
remains unchanged
0%
decreases
0%
none of these

Explanation

As capacitance,

$C \propto k$ where

$k=$ dielectric constant.
and potential

$V \propto \frac{1}{k}$ ,
Energy

$U=\frac{1}{2}CV^2 \Rightarrow U \propto \frac{1}{k}$
Thus when dielectric present , energy will decrease and when it is removed ,the energy will increase.

When dielectric medium of constant k is filled between the plates of a charged parallel-plate condenser, then the energy stored becomes, as compared to its previous value,

0%
$K^{-3} times$
0%
$K^{-2} times$
0%
$K^{-1} times$
0%
$K times$

Explanation

As no cell is connected to the capacitor so the charge (Q) remains constant.
If the capacitance of capacitor before inserting the dielectric is

$C$ then capacitance after inserting dielectric becomes

$KC$
Before inserting dielectric the energy is

$U=\dfrac{Q^2}{2C}$
After inserting dielectric the energy is

$U'=\dfrac{Q^2}{2KC}=UK^{-1}$

Which of the following units is not equivalent to Farad?

0%
$CV^2$
0%
$J/V^2$
0%
$Q^2/J$
0%
$Q/V$

Explanation

Farad is the S I unit of capacitance. The conversion units of Farad are

$F = \dfrac {A.s}{V} = \dfrac {J} {V^2} = \dfrac {W.s}{V^2} = \dfrac {C} {V} = \dfrac {C^2}{J}$
where

$A$ is ampere,

$V$ is volt,

$C$ is coulomb,

$W$ is watt and

$J$ is joule.

The capacitance of a condenser is

$20 \mu F$ and it is charged to a potential of 2000 V. The energy stored in it will be :

0%
zero
0%
40 J
0%
80 J
0%
120 J

Explanation

Given

$C=20 \mu F, V=2000 V$
Energy stored in the condenser is

$U=\frac{1}{2}CV^2=\frac{1}{2}\times 20\times10^{-6}\times (2000)^2=40 J$

A capacitor of capacitance C is connected to battery of emf

$V_0$ . Without removing the battery, a dielectric of strength

$\varepsilon _r$ is inserted between the parallel plates of the capacitor C, then the charge on the capacitor is :

0%
$CV_0$
0%
$\varepsilon _rCV_0$
0%
$\dfrac{CV_0}{\varepsilon _r}$
0%
none of these

Explanation

As the battery is not removed so the potential remains constant and it is equal to

$V_0$ .
Due to insert the dielectric the capacitance

$C$ will become

$C'=\varepsilon_r C$
Thus the charge on the capacitor

$Q=C'V_0=\varepsilon_r CV_0$

The net charge on a condenser is :

0%
infinity
0%
q/2
0%
2q
0%
zero

Explanation

Every condenser is made with two plates. The charge on one plate is

$+Q$ and other is

$-Q$ . Thus total charge of condenser is

$Q_t=+Q-Q=0$

A parallel plate condenser with plate separation d is charged with the help of battery so that

$V_0$ energy is stored in the system. The battery is now removed. a plate of dielectric constant k and thickness d is placed between the plates of condenser. The new energy of the system will be :

0%
$V_0 k^{-2}$
0%
$k^2 V_0$
0%
$V_0 k^{-1}$
0%
$k V_0$

Explanation

As the battery is removed from the capacitor so the charge (Q) remains constant.
If the capacitance of capacitor before inserting the dielectric is

$C$ , the capacitance of capacitor after inserting the dielectric is

$kC$
Here, Energy

$V_0=\dfrac{Q^2}{2C}$
After inserting dielectric the energy becomes

$U=\dfrac{Q^2}{2kC}=V_0k^{-1}$

The energy stored between the plates of a condenser in not represented by :

0%
$U=\dfrac{CV^2}{2}$
0%
$U=2qV$
0%
$U=\dfrac{q^2}{2C}$
0%
$U=\dfrac{qV}{2}$

Explanation

Energy stored in capacitor

$U=\dfrac{1}{2}CV^2$
Since,

$q=CV$ ,

$U=\dfrac{1}{2}CV.V=\dfrac{1}{2}qV$
when

$V=\dfrac{q}{C}$ ,

$U=\dfrac{1}{2}C(\dfrac{q}{C})^2=\dfrac{q^2}{2C}$
thus

$U=2qV$ is not the expression of energy stored in capacitor.

Which of the following is correct statement :

0%
Equipotential lines are always perpendicular to the electric field
0%
Work done for moving a charge along the conducting surface (closed and containing charge) very close to it may be negative or positive
0%
Electric field may cross each other
0%
None of the above

A battery of 100 V is connected to series combination of two identical parallel-plate condensers. If dielectric of constant 4 is slipped between the plates of second condenser, then the potential difference on the condensers will respectively become :

0%
80 V, 20 V
0%
75 V, 25 V
0%
50 V, 80 V
0%
20 V, 80 V

Explanation

Let the capacitance of each capacitors before applied dielectric is

$C$ .

When dielectric is inserted in the second capacitor , the capacitance of first capacitor will remain same i.e

$C_1=C$ and the capacitance of second capacitor becomes

$C_2=4C$

As they are in series, equivalent capacitance,

$C_{eq}=\dfrac{C_1C_2}{C_1+C_2}=\dfrac{C. 4C}{C+4C}=\dfrac{4}{5}C$

Equivalent charge on the circuit is

$Q_{eq}=C_{eq}V=\dfrac{4}{5}C\times 100=80C$

Now the potential difference on first condenser is

$V_1=\dfrac{Q_{eq}}{C_1}=\dfrac{80C}{C}=80 V$

Now the potential difference on second condenser is

$V_2=\dfrac{Q_{eq}}{C_2}=\dfrac{80C}{4C}=20 V$

The energy stored between the plates of a condenser in not represented by :

0%
$U=\dfrac{CV^2}{2}$
0%
$U=2qV$
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$U=\dfrac{q^2}{2C}$
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$U=\dfrac{qV}{2}$

Explanation

Energy stored in capacitor

$U=\dfrac{1}{2}CV^2$
Since,

$q=CV$ ,

$U=\dfrac{1}{2}CV.V=\dfrac{1}{2}qV$
when

$V=\dfrac{q}{C}$ ,

$U=\dfrac{1}{2}C(\dfrac{q}{C})^2=\dfrac{q^2}{2C}$
thus

$U=2qV$ is not the expression of energy stored in capacitor.

The capacitance of a charged condenser is C and energy stored on account of charge on it is U, then the quantity of charge on the condenser will be :

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$\sqrt{2UC}$
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$\sqrt{\frac{UC}{2}}$
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$2UC$
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zero

Explanation

The energy stored in the condenser is

$U=\dfrac{q^2}{2C}$
so the charge on the condenser is

$q=\sqrt{2UC}$

The energy acquired by a charged particle of

$4 \mu C$ when it is accelerated through a potential difference of 8 Volt will be :

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$3.2\times 10^{-7} J$
0%
$3.2\times 10^{-5} J$
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$2\times 10^{-6} J$
0%
$2\times 10^{-5} J$

Explanation

One volt is the potential difference between two points in an electric field such that one joule of work is done in moving one coulomb of charge from one point to another.

$W=QV$

$U=4\times 10^{-6}\times 8=32\times 10^{-6}=3.2\times 10^{-5} J$

Three condensers each of capacitance 2 F, are connected in series. The resultant capacitance will be :

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6 F
0%
5 F
0%
2/3 F
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3/2 F

Explanation

Let the resultant capacitor is

$C_{R}$
For series combination of three capacitors ,

$\dfrac{1}{C_R}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}$ F

$\therefore C_R=\dfrac{2}{3}F$

When two condensers of capacitance

$1\mu F$ and

$2\mu F$ are connected is series then the effective capacitance will be :

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$\dfrac{2}{3}\mu F$
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$\dfrac{3}{2}\mu F$
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$3\mu F$
0%
$4\mu F$

Explanation

When two condenser are in series , the equivalent capacitance

$C_{eq}=\dfrac{C_1C_2}{C_1+C_2}=\dfrac{1\times2}{1+2}=\dfrac{2}{3} \mu F$

What will be area of pieces of paper in order to make a paper condenser of capacitance

$0.04 \mu F$ , if the dielectric constant of paper is

$2.5$ and its thickness is

$0.025 mm$ ?

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$1m^2$
0%
$2\times 10^{-3} m^2$
0%
$4.51\times10^{3} m^2$
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$10^{-3} m^2$

Explanation

For paper condenser we will use the expression of capacitance of parallel plate condenser with dielectric.

$C=\dfrac{Ak\epsilon_0}{d}$

$A=\dfrac{Cd}{k\epsilon_0}=\dfrac{0.04\times 10^{-6}\times 0.025\times 10^{-3}}{2.5\times 8.854\times 10^{-12}}=4.51\times 10^{3} m^2$

In a charged capacitor, the energy resides in :

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the positive charges
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both the positive and negative charges
0%
the field between the plates
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around the edge of the capacitor plates

The capacitance of a parallel plate capacitor in air is

$2\ \mu F$ . If a dielectric medium is placed between the plates then the potential difference reduces to

$\dfrac{1}{6}$ of the original value. The dielectric constant of the medium is :

0%
$6$
0%
$3$
0%
$2.2$
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$4.4$

Explanation

As no battery is connected to the capacitor, the charge remains constant.
Before inserting the dielectric, the charge is

$Q=CV$
After inserting the dielectric, the capacitance become

$C'=KC$ and hence charge

$Q'=\dfrac{C'V}{6}=\dfrac{KCV}{6}$
As the charge is constant ,

$Q=Q'$

$\Rightarrow CV=\dfrac{KCV}{6}$

$\therefore K=6$

A slab X is placed between the two parallel isolated charged plates as shown in the figure. If

$E_p$ and

$E_q$ denotes the intensity of electric field at P and Q, then :

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$E_p$ is reduced by the presence of X, if X is metallic
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$E_q$ is increased by presence of X, if X is dielectric
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$E_q$ is in the opposite sense to $E_p$ , if X is dielectric
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$E_q$ is zero, if X is metallic

Explanation

There is no effect of the metal on the external electric field(thus A is incorrect) while the dielectric reduces the net electric field outside. The dielectric produces an electric field inside it due to the induced charges which is opposite to the external field. Thus

$E_Q$ is reduced and B and C are also incorrect.(Note that

$E_P$ is not the external field). Now as the electric field inside the conductor is zero, the field

$E_p$ is zero if X is a metallic. Thus D is correct.

The capacitance of parallel-plate capacitor is

$4\mu F$ . If a dielectric material of dielectric constant 16 is placed between the plates then the new capacitance will be :

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$1/64 \mu F$
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$0.25 \mu F$
0%
$64 \mu F$
0%
$40 \mu F$

Explanation

If the capacitance of the parallel plate capacitor without dielectric is

$C$ , the capacitance of it with dielectric will be

$C'=kC=16\times 4=64 \mu F$

A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If

$'x'$ is the separation between the plates, then the time rate of change of electrostatic energy of the capacitor is proportional to :

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$x^{2}$
0%
$x$
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$x^{-1}$
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$x^{-2}$

Explanation

Parallel plate capacitance at instant time t ,

$C=\dfrac{\epsilon_0 A}{x}$

Also potential

$(V)$ will remain same (constant) because battery is attached.

Energy

$U=\dfrac{1}{2}CV^2=\dfrac{1}{2}\times \dfrac{\epsilon_0 A}{x}\times V^2$

$U = \dfrac{1}{2} \epsilon_0 A V^2(\dfrac{1}{x})$

Thus,

$U \propto \dfrac{1}{x}$

$\dfrac{dU}{dt} = \dfrac{1}{2} \epsilon_0 A V^2 (-\dfrac{1}{x^2} \dfrac{dx}{dt})$

$\dfrac{dU}{dt} \propto \dfrac{1}{x^2}$

Thus rate of change of U is proportional to

$x^{-2}$ .

A condenser is charged to a potential difference of 200 volts as a result of which it gains charge of 0.1 coulomb. When it is charged then the energy released will be :

0%
1 J
0%
2 J
0%
10 J
0%
20 J

Explanation

The expression of energy released is

$U=\dfrac{1}{2}CV^2$
as

$Q=CV$ thus,

$U=\dfrac{1}{2}CV.V=\dfrac{1}{2}QV=\dfrac{1}{2}\times 0.1\times 200=10 J$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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