MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 14

The equivalent capacitance between the points

$A$ and

$B$ in the given diagram is:

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$8\mu F$
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$6\mu F$
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$\frac{8}{3}\mu F$
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$\frac{3}{8}\mu F$

A non conducting semicircular disc (as shown in figure) has a uniform surface charge density

$\sigma$ . The electric potential at the centre of the disc:-

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$\dfrac{\sigma \ \iota n(b/a)}{2\pi \ \varepsilon _{0}\ (b-a)}$
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$\dfrac{\sigma (b-a)}{2\ \varepsilon _{0}}$
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$\dfrac{\sigma (b-a)}{4\ \varepsilon _{0}}$
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$\dfrac{\sigma (b-a)}{4\pi \ \varepsilon _{0}}$

A sliding rod AB of resistance R is shown in the figure. Here magnetic field B is constant and is Out of the paper. Parallel wires have no resistance and the rod is moving with Constant velocity v. The current in the sliding rod AB in the function of t, when switch S is closed at time t = 0 is

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$\left(\dfrac{vBd}{R}\right )e^{-t/C}$
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$\left(\dfrac {vBd}{R} \right)e^{-t/RC}$
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$\left(\dfrac {vBd}{R}\right )e^{RtC}$
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$\left(\dfrac {vBd}{R}\right )e^{t/RC}$

Explanation

Here magnetic field

$\vec{B}$ is constant and is out of paper.
When the sliding rod AB moves with a velocity v in the direction shown in the figure, the induced current in AB is from A to B.
As the switch S is closed at time t = 0, the capacitor gets charged. If q is the charge on the capacitor, then

$I \, = \, \dfrac{dq}{dt} \, = \, \dfrac{Bdv}{R} \, - \,\dfrac{q}{RC} \,\,\,\,\, or \,\,\,\,\, \dfrac{q}{RC} + \dfrac{dq}{dt} \, = \, \dfrac{Bdv}{R}$

$q \, = \, vBdC + A\, e^{t/RC}$ (where A is a constant)

At t = 0, q = 0

$\therefore A = -vBdC$
From

$(i) q = vBdC [ 1 - e^{-t/RC}]$

$\displaystyle I \, = \, \dfrac{dq}{dt} \, = \, vBdC \times \frac{1}{RC}e^{-t/RC} \, = \, \frac{vBd}{R}e^{-t/RC}$

1047336_942497_ans_4878f32c07a047dbae6a7caa5cccc1fb.png

In a quark model of elemetary particles, a neutron is made of one up quark of charge

$\cfrac{2}{3}e$ and two down quark of charges

$\left( -\cfrac { 1 }{ 3 } e \right)$ . If they have a triangle configuration with side length of the order of

${ 10 }^{ -15 }m$ . The electrostatic potential energy of neutron in

$MeV$ is

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$7.68$
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$-5.21$
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$-.048$
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$9.34$

Explanation

Fig shows the quark model of a neutral, where

$r=10^{-15}m$ .

Potential energy of neutron

$V=\dfrac{1}{4\pi \epsilon_0 r}[q_eq_d+q_uq_d+q_uq_d]$

$=\dfrac{9\times 10^9}{10^{-15}}\left[\left(-\dfrac{e}{3}\right)\left(-\dfrac{e}{3}\right)+\left(\dfrac{2e}{3}\right)\left(-\dfrac{e}{3}\right)+\left(\dfrac{2e}{3}\right)\left(-\dfrac{e}{3}\right)\right]$

$V=\dfrac{9\times 10^9}{10^{-15}}\left(\dfrac{1}{9}-\dfrac{4}{9}\right)(1.6\times 10^{-19})^2=-7.68\times 10^{-14}J$

$V=\dfrac{-7.68\times 10^{-14}}{1.6\times 10^{-19}}eV=-4.8\times 10^5eV=-0.48MeV$

The gravitational field strength

$\vec {E}$ and gravitational potential

$V$ are related as

$\vec {E} = -\left (\dfrac {\partial V}{\partial x} \hat {i} + \dfrac {\partial V}{\partial y} \hat {j} + \dfrac {\partial V}{\partial z}\hat {k}\right )$
In the figure, transversal lines represent equipotential surfaces. A particle of mass

$m$ is released from rest at the origin. The gravitational unit of potential,

$1\overline {V} = 1\ cm^{2}/ s^{2}$ .
X-component of the velocity of the particle at the point

$(4cm, 4cm)$ is

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$$4\ cm/s$
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$2\ cm/s$
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$2\sqrt {2} \dfrac {cm}{s}$
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$1\ cm/s$

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 the battery still in connection. The corresponding quantities now given by

$Q, V, E$ and

$U$ are related to the previous one as

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$Q>{ Q }_{ 0 }$
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$V>{ V }_{ 0 }$
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$E>{ E }_{ 0 }$
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$U>{ U }_{ 0 }$

A parallel plate capacitor of plate area

$A$ and plate separation

$d$ is charged to potential difference

$V$ and then the battery is disconnected. A slab of dielectric constant

$K$ is then inserted between the plate of the capacitor so as to fill the space between the plates. If

$Q, E$ and

$W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab inserted), and work done on the system, in question, in the process of inserting the slab, then

0%
$Q=\dfrac { { \varepsilon }_{ 0 }AV }{ d }$
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$Q=\dfrac { { \varepsilon }_{ 0 }KAV }{ d }$
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$E=\dfrac { V }{ Kd }$
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$W=\dfrac { { \varepsilon }_{ 0 }{ AV }^{ 2 } }{ 2d } \left[ 1-\dfrac { 1 }{ K } \right]$

Two point charges placed 3m apart, have a combined charge 20

$\mu$ C. Find the two charges if they repeleach other with a force 0.075 N.

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25 $\mu$ C, -5 $\mu$ C
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30 $\mu$ C, 10 $\mu$ C
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30 $\mu$ C, 50 $\mu$ C
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15 $\mu$ C, 5 $\mu$ C

A capacitor of $2F$ is charged as shown in the diagram. When the switch $S$ is tuned to position

$2$ , the percentage of its stored energy dissipated is :

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$80$ %
0%
$0$ %
0%
$20$ %
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$75$ %

Explanation

$U_{at\ 1}= \dfrac{1}{2} \times 2 \times v^2 = v^2$

When S at 2,

Consider loop as a series:

$C_{eq} = 10 \mu F$

$V_eq= \dfrac{Q}{C_eq} \, = \dfrac{2\times V}{C_eq}$

$V_{eq} = \dfrac{2V}{2 + 8} = \dfrac{V}{5}$

$U_{at} 2 = \dfrac{1}{2} \times 10 \times \left(\dfrac{V}{5} \right)^2 = \dfrac{V^2}{5}$

$\Delta U = \dfrac{4V^2}{5}$ . i.e energy loss

% loss

$= \dfrac{4V^2}{5V^2} \times 100 = 80 \%$ Loss

Option A is correct.

1450433_1026361_ans_12e27009b9804943bdd37f71083b56bb.png

The gravitational field strength

$\vec {E}$ and gravitational potential

$V$ are related as

$m$ is released from rest at the origin. The gravitational unit of potential,

$1\overline {V} = 1\ cm^{2}/ s^{2}$ .
Speed of the particle

$(v) (y$ is in cm

$v$ is in

$cm/s$ ) as function of its y-co-ordinate is

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$v = 2\sqrt {y}$
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$v = \sqrt {2}. y$
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$v = 2y$
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$v = 2y + 4y^{2} + 2y^{2}$

One plate of a parallel plate capacitor is suspended from a beam of a physical balance as shown in figure. The area of each plate is

$625 \ cm^2$ and the distance between these plates is

$5 \ mm$ . If an additional mass

$0.04 \ gm$ is placed in the other pan of the balance, then the potential difference required between the plates to keep it in equilibrium will be:

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$150 \ V$
0%
$188 \ V$
0%
$225 \ V$
0%
$310 \ V$

Explanation

One plate capacitor

parallel capacitor

The area of each plate

$=625cm^2$

Distance between these plate

$=5mm$

Additional mass

$=0.04gm$

$C=\cfrac{E_0A}{d}\\ \quad=\cfrac{8.854\times(625\times10^{-4})}{5\times10^{-3}}\\ \quad=110.675\\V=Q/C=\cfrac{0.04\times10^{-3}}{110.675}=225V$

1231379_1065915_ans_73261f137978486b929857e5acb85a12.png

The diagram shows a small bead of mass

$m$ carrying charge

$q$ . The bead can freely move on the smooth fixed ring placed on a smooth horizontal plan. In the same plane a charge

$+Q$ has also been fixed as shown. The potential at the point

$P$ due to

$+Q$ is

$V$ . The velocity with which the bead should projected from the point

$P$ so that it can complete a circle should be greater than

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$\sqrt{\dfrac{6qV}{m}}$
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$\sqrt{\dfrac{qV}{m}}$
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$\sqrt{\dfrac{3qV}{m}}$
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$none$

Explanation

Hint:

Potential due to charge $q$ at distance $r$ from it is given as,

$V = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q}{r}$

Potential Energy of charges $q_1$ and $q_2$ at distance $r$ is given as,

$U = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q_1 q_2}{r}$

Correct Option: A.

Step 1: Note all the values given and find the condition for which charge can complete the circle.

Mass of small charge on the bead is

$m$ , and charge on it is

$+q$ , it is placed on bead such that it moves on ring smoothly. That is, there is no energy loss by friction.

The charge

$+Q$ is fixed in the ring as given in the figure.

Charge

$+q$ and

$+Q$ both are positive, thus they repel each other. If we somehow bring the charge to the point which is diametrically opposite to point P, the repulsion Force on

$+q$ will increase and it will itself move back to point P thus completing the circle. So we only need to move the charge to complete semicircle.

Step 2: Find the change in Potential Energy.

Potential due to charge

$+Q$ at point P is given as,

$V = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{4a} = qv$

Potential Energy of charges when

$+q$ is at point P is givne as,

$U_i = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Qq}{4a} = qv$

Potential Energy of charges when

$+q$ is at point which is diametrically opposite to point P, is

$U_f = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Qq}{a}$

$U_f = 4qV$

Step 3: Find the velocity of the bead with which it should be projected.

Let us project the bead with velocity

$v$ , and let it stop at the point when it completes semicircle. Since, no external forces are present, by conservation of energy, we have

$-(U_f - U_i) = KE_f - KE_i$

$\Rightarrow -(4qV - qV) = 0 - \dfrac{1}{2}mv^2$

$\Rightarrow 3qV = \dfrac{1}{2}mv^2$

$\Rightarrow \dfrac{6qV}{m} = v^2$

$\Rightarrow v = \sqrt{\dfrac{6qV}{m}}$

Thus, minimum velocity with which bead should be projected is given as, $v = \sqrt{\dfrac{6qV}{m}}$ .

A uniformly charged solid sphere of radius

$R$ has potential

$V_{0}$ (measured with respect to

$\infty$ ) on its surface. For this sphere the equipotential surfaces with potentials

$\dfrac {3V_{0}}{2}, \dfrac {5V_{0}}{4}, \dfrac {3V_{0}}{4}$ and

$\dfrac {V_{0}}{4}$ have radius

$R_{1}, R_{2}, R_{3}$ , and

$R_{4}$ respectively. Then

0%
$R_{1} = 0$ and $R_{2} > (R_{4}\ R_{3})$
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$R_{1} \neq 0$ and $(R_{2}\ R_{1}) > (R_{4}\ R_{3})$
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R $_{1} = 0$ and $R_{2} < (R_{4}\ R_{3})$
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$2R < R_{4}$

Explanation

Radius

$=R$

Potential

$={ V }_{ 0 }$

with potentials

$=\dfrac { { 3V }_{ 0 } }{ 2 } ,\dfrac { 5{ V }_{ 0 } }{ 4 } ,\dfrac { 3{ V }_{ 0 } }{ 4 } ,\dfrac { { V }_{ 0 } }{ 4 }$

with radius

$={ R }_{ 1 },{ R }_{ 2 },{ R }_{ 3 },{ R }_{ 4 }$ respectively.

$(C)$ option will be correct.

The potential at the surface of the charged sphere,

${ V }_{ \phi }=\dfrac { KQ }{ R }$

$V=\dfrac { KQ }{ r } \times r\ge R$

$=\dfrac { KQ }{ { 2R }^{ 3 } } \left( { 3R }^{ 2 }-{ r }^{ 2 } \right)$ ;

$r\le R$

${ V }_{ centre }={ V }_{ C }=\dfrac { KQ }{ 2{ R }^{ 3 } } \times { 3R }^{ 2 }=\dfrac { 3 }{ 2 } \times KQ=\dfrac { { 3V }_{ 0 } }{ 2 }$

As potential decreases for outside points. Thus, according to the question, we can write

${ V }_{ { R }_{ 2 } }=\dfrac { 5{ V }_{ 0 } }{ 4 } =\dfrac { KQ }{ 2{ R }^{ 3 } } \left( { 3R }^{ 2 }-{ R }_{ 2 }^{ 2 } \right)$

$\dfrac { { 5V }_{ 0 } }{ 4 } =\dfrac { { V }_{ 0 } }{ 2{ R }^{ 2 } } \left( { 3R }^{ 2 }-{ R }_{ 2 }^{ 2 } \right)$

or,

$\dfrac { 5 }{ 2 } =3-{ \left( \dfrac { { R }_{ 2 } }{ R } \right) }^{ 2 }$

or,

${ \left( \dfrac { { R }_{ 2 } }{ R } \right) }^{ 2 }=3-\dfrac { 5 }{ 2 } =\dfrac { 1 }{ 2 }$

or,

${ R }_{ 2 }=\dfrac { R }{ \sqrt { 2 } }$ ,

Similarly,

${ V }_{ { R }_{ 3 } }=\dfrac { { 3V }_{ 0 } }{ 4 }$

or,

$\dfrac { KQ }{ { R }_{ 3 } } =\dfrac { 3 }{ 4 } \times \dfrac { KQ }{ R }$

or,

${ R }_{ 3 }=\dfrac { 4 }{ 3 } R$

or,

${ V }_{ { R }_{ 4 } }=\dfrac { KQ }{ { R }_{ 4 } } \times \dfrac { { V }_{ 0 } }{ 4 }$

or,

$\dfrac { KQ }{ { R }_{ 4 } } =\dfrac { 1 }{ 4 } \times \dfrac { KQ }{ R }$

or,

${ R }_{ 4 }=4R$

Now,

${ R }_{ 1 }=0$ and

${ R }_{ 2 }<\left( { R }_{ 4 }{ R }_{ 3 } \right)$

1244650_1026104_ans_540c9ccd313a4826b090e34ed7093333.jpg

A dipole of

$2 \mu C$ charges each, consists of the positive charge at the point

$P(1, -1)$ and the negative charge is placed at the point

$Q(-1, 1)$ . The work done in displacing a charge of

$+1 \mu C$ from point

$A(-3, -3)$ to

$B(4, 4)$ is:

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$1.6 \times 10^{-19}J$
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$3.2 \times 10^{-19}J$
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zero
0%
$4.8 eV$

Explanation

The charge

$1\mu C$ is moved along equitorial line of the dipole. Potential of dipole along equitorial line is zero. Hence, work done in displacing

$1\mu C$ along AB is zero.

1102778_1016459_ans_09bfbaba8e5f4fecb16903ba81c39d68.png

Two equal charges

$q$ are placed at a distance of

$2a$ and a third charge

$-2q$ is placed at the midpoint. The potential energy of the system is

0%
$\dfrac { { q }^{ 2 } }{ 8\pi { \varepsilon }_{ 0 }a }$
0%
$\dfrac { 6{ q }^{ 2 } }{ 8\pi { \varepsilon }_{ 0 }a }$
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$-\dfrac { 7{ q }^{ 2 } }{ 8\pi { \varepsilon }_{ 0 }a }$
0%
$\dfrac { 9{ q }^{ 2 } }{ 8\pi { \varepsilon }_{ 0 }a }$

The variation of potential with distance

$R$ from a fixed point is as shown below. A charge

$q$ is put at

$R=3\ m$ and released. The charge will

0%
Move away from origin
0%
Move towards the origin
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Not move
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Trace a circular path

The electric potential at a point in free space due to a charge

$Q$ coulomb is

$Q\times {10}^{11}V$ . The electric field at that point is

0%
$4\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 22 }V/m$
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$127\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 20 }V/m$
0%
$4\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 20 }V/m$
0%
$12\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 22 }V/m$

Explanation

$V=Q\times 10^{11}$

$E=?$

$E=\dfrac{V}{r}=\frac{Q}{4\pi \varepsilon_0 r^2}$

$V=\dfrac{Q}{4\pi \varepsilon_0 r}=Q\times 10^{11}$

$r=\dfrac{Q}{4\pi \varepsilon_0\times Q\times 10^{11}}$

$r=\dfrac{1}{4\pi \varepsilon_0\times 10^{11}}$

$E=\dfrac{V}{r}=\dfrac{Q\times 10^{11}}{\dfrac{1}{4\pi \varepsilon_0\times 10^{11}}}=4\pi \varepsilon_0Q\times 10^{22} V/m$

The distance between electric charges

$1\mu C$ and

$3\mu C$ is

$8m$ . What is the electric potential at a point on the line joining them where the electric field is zero? (

$k=9\times {10}^{9}$ SI)

0%
$9\times {10}^{2}$
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$9\times {10}^{3}$
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$9\times {10}^{4}$
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$9\times {10}^{5}$

A dielectric slab of thickness

$d$ inserted in a parallel plate capacitor whose negative plate is at

$x=0$ and positive plate is at

$x=3d$ . The slab is equidistant from the plates. The capacitor is given some charge .As

$x$ goes from

$0$ to

$3d$ :

0%
$The\ magnitude\ of\ the\ elecrtric\ field\ remains\ the\ same.$
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$The\ direction\ of\ the\ electric\ field\ remains\ the\ same.$
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$The\ electric\ potential\ increases\ continuously.$
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$The\ electric\ potential\ increase\ at\ first\, then\ decreases\ and\ again\ increases.$

The capacity of a parallel plate capacitor formed by the plates of same are A is 0.02

$\mu F$ with air as dielectric. Now one plate is replaced by a plate of area 2A and dielectric (K = 2) is introduced between the plates, the capacity is :

0%
0.04 $\mu F$
0%
0.08 $\mu F$
0%
9.91 $\mu F$
0%
2 $\mu F$

$4\ \mu F$ and

$6\ \mu F$ capacitors are joined in series and

$500\ v$ are applied between the outer plates of the system. What is the charge on each plate ?

0%
$1\cdot 2\times 10^{3}\ C$
0%
$6\cdot 0\times 10^{3}\ C$
0%
$5\cdot 0\times 10^{-3}\ C$
0%
$2\cdot 0\times 10^{-3}\ C$

Explanation

Since the capacities are connected in series

$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$

$\dfrac{1}{C_{eq}}=\dfrac{1}{4\mu}+\dfrac{1}{6\mu}$

$C_{eq}=\dfrac{12\mu}{5}$

$C=\dfrac{Q}{V}$

$\dfrac{12\mu}{5}=\dfrac{Q}{500}$

$Q=1.2\times 10^{3}$

Select the correct statements:

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The electric lines of force are always closed curves.
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Electric line of forced is parallel to equipotential surface
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Electric line of force is perpendicular to equipotential surface.
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Electric line of force is always the path of a positively charged particle.

Three long concentric cylindrical shells have radii R, 2R and

$2\sqrt{2}R$ . Inner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is:

0%
$\dfrac{\dfrac{1}{3}\epsilon_0}{ln 2}$
0%
$\dfrac{6\pi \epsilon_0}{ln 2}$
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$\dfrac{\pi \epsilon_0}{2 ln 2}$
0%
None

The capacitance of a capacitor is

$10F$ . The potential difference on it is

$50V$ . If the distance between its plate is halved, What will be the potential difference now?

0%
$100V$
0%
$50V$
0%
$25V$
0%
$75V$

Explanation

Given,

$C=10F$

$V=50V$
If the distance between the plate is halved.

$C=\cfrac{A\epsilon_0}{d}$

$C^{\prime}=\cfrac{2A\epsilon_0}{d}=2C$
Since energy is not dissipated, it remain constant.
The surface charge density

$(\sigma)$ will remain constant.

$E=\cfrac{\sigma}{\epsilon_0}\quad$ [

$\therefore E$ is also constant]

$V=E\times d$

$\Rightarrow V^{\prime}=Ed^{\prime}$

$\Rightarrow V^{\prime}=\cfrac{V}{2}=\cfrac{50}{2}=25$
This shows that V will also get halved.

Four condenser each of capacity

$4\mu F$ are connected as shown in figure

${V_p} - {V_q} = 15$ volts. The energy stored in the system is

0%
$2400$ ergs
0%
$1800$ ergs
0%
$3600$ ergs
0%
$5400$ ergs

Explanation

The resultant capacitance is,

$\dfrac{1}{C} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{4} = \dfrac{5}{8}$

$C = \dfrac{8}{5} = 1.6{\rm{\mu F}}$

Energy stored in capacitor is,

$E= \dfrac{1}{2}C{V^2}$

$= \dfrac{1}{2} \times 1.6 \times {10^{ - 6}} \times {15^2}$

$= 0.8 \times 225 \times {10^{ - 6}}{\rm{J}}$

$= 1800\;{\rm{erg}}$

Two condensers of capacity

$0.3 \mu F$ and

$0.6 \mu F$ respectively are connected in series. The combination is connected across a potential of

$6$ volts. The ratio of energies stored by the condensers will be:

0%
$\dfrac {1}{2}$
0%
$2$
0%
$\dfrac {1}{4}$
0%
$4$

Explanation

Since the capacitors are connected in series so charge flowing through them will be same.

Energy stored in first capacitor of capacitance $0.3\;{\rm{\mu F}}$ is given by

${E_1} = \frac{{{Q^2}}}{{2{C_1}}}$

${E_1} = \frac{{{Q^2}}}{{2 \times 0.3}}$

Energy stored in second capacitor of capacitance $0.6\;{\rm{\mu F}}$ is given by

${E_2} = \frac{{{Q^2}}}{{2{C_2}}}$

${E_2} = \frac{{{Q^2}}}{{2 \times 0.6}}$

Now ratio of energy stored by the condensers is given by

$\frac{{{E_1}}}{{{E_2}}} = \frac{2}{1}$

Two materials of dielectric constants

$k_1\, and \, k_2$ are introduced to fill the space between the two parallel plates of a capacitor as shown in the figure. The capacitance of the capacitor is :

0%
$\dfrac{{A{ \in _0}\left( {{k_1} + {k_2}} \right)}}{{2d}}$
0%
$\dfrac{{2A{ \in _0}}}{d}\left( {\dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}} \right)$
0%
$\dfrac{{A{ \in _0}}}{d}\left( {\dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}} \right)$
0%
$\dfrac{{A{ \in _0}\left( {{k_1} + {k_2}} \right)}}{{2d{k_1}{k_2}}}$

Explanation

$C=K\dfrac { { E }_{ 0 }A }{ d }$

$C={ C }_{ 1 }+{ C }_{ 2 }=\dfrac { { E }_{ 0 }\times \dfrac { A }{ 2 } \times { K }_{ 1 } }{ d } +\dfrac { { E }_{ 0 }\times \dfrac { A }{ 2 } \times { K }_{ 2 } }{ d } =\dfrac { { E }_{ 0 }A\left( { K }_{ 1 }+{ K }_{ 2 } \right) }{ 2d }$

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness

$d_{1}$ and dielectric constant

$k_{1}$ and the other has thickness

$d_{2}$ and dielectric constant

$k_{2}$ as shown in fig. This arrangement can be thought as a dielectric slab of thickness

$d=d_{1}+d_{2}$ and effective dielectric constant

$k$ . The

$k$ is then :

0%
$\dfrac{k_{1}d_{1}+k_{2}d_{2}}{d_{1}+d_{2}}$
0%
$\dfrac{k_{1}d_{1}+k_{2}d_{2}}{k_{1}+k_{2}}$
0%
$\dfrac{k_{1}k_{2}(d_{1}+d_{2})}{k_{1}d_{2}+k_{2}d_{1}}$
0%
$\dfrac{2k_{1}k_{2}}{k_{1}+k_{2}}$

Explanation

For the first block

Thickness $={{d}_{1}}$

Dielectric constant $={{k}_{1}}$

Capacitance $={{C}_{1}}$

For second block

Thickness $={{d}_{2}}$

Dielectric constant $={{k}_{2}}$

Capacitance $={{C}_{2}}$

Capacitors C₁ and C₂ are connected in parallel. So equivalent capacitance is :
$\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$

${{C}_{eq}}=\dfrac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}........(1)$

Capacitance in parallel plate capacitor is given as

${{C}_{1}}=\dfrac{{{k}_{1}}{{\varepsilon }_{0}}A}{{{d}_{1}}}\,\,\And \,\,{{C}_{2}}=\dfrac{{{k}_{2}}{{\varepsilon }_{0}}A}{{{d}_{2}}}$

So, equation (1) becomes

${{C}_{eq}}=\dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}.............(2)$

Equivalent capacitance for the combination is

$C=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}}.........(2)$

From equation (1) and (2)

$\dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}}$

$k=\dfrac{{{k}_{1}}{{k}_{2}}({{d}_{1}}+{{d}_{2}})}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}$

Electrical potential in an electric field is given by V=K/r (K being constant). If the position vector

$\underset{r}{\rightarrow}=2\widehat{i}+3\widehat{j}+6\widehat{k}$ , then the electric field will be

0%
$\dfrac{2\widehat{i}+3\widehat{j}+6\widehat{k}}{243}K$
0%
$\dfrac{2\widehat{i}+3\widehat{j}+6\widehat{k}}{343}K$
0%
$\dfrac{3\widehat{i}+2\widehat{j}+6\widehat{k}}{243}K$
0%
$\dfrac{3\widehat{i}+2\widehat{j}+6\widehat{k}}{343}K$

Capacitance

$C_1 = 2 C_2 = 2C_3$ and potential difference across

$C_1, C_2$ and

$C_3$ are

$V_1, V_2$ and

$V_3$ respectively then:

0%
$V_1 = V_2 = V_3$
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$V_1 = 2V_2 = 2V_3$
0%
$2V_1 = V_2 = V_3$
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$2V_1 = 2V_2 = V_3$

A parallel-plate capacitor of plate area

$A$ and plate separation

$d$ is charged to a potential difference and then the battery is disconnected. A slab of dielectric constant

$K$ is then inserted between the plates of the capacitor so as to fill the whole space between the plates. Find the work done on the system the process of inserting the slab.

0%
$\dfrac{\varepsilon_0AV^2}{2d}\left({{1}-{\dfrac{1}{K}}}\right)$
0%
$\dfrac{\varepsilon_0AV^2}{d}\left(\dfrac{1}{K}-1\right)$
0%
$\dfrac{\varepsilon_0AV^2}{2d}\left(\dfrac{1}{K}+1\right)$
0%
$\dfrac{\varepsilon_0AV^2}{d}\left(\dfrac{1}{K}+1\right)$

Explanation

after inserting slab

$new\quad capacitance { C }^{ \prime }=KC=\dfrac { K{ \epsilon_o }A }{ d }$

$new\quad potential\quad difference{ V }^{ \prime }=\frac { V }{ K }$

$new\quad electricfield\quad E^{ \prime }=\dfrac { { V }^{ \prime } }{ d } =\dfrac { V }{ kd }$

$new\quad charge{ Q }^{ \prime }{ C }^{ \prime }=\dfrac { { \epsilon_o }AV }{ d }$

$work=final-initial$

$=\dfrac { 1 }{ 2 } { C }^{ \prime }{ V }^{ \prime }-\dfrac { 1 }{ 2 } { CV }^{ 2 }$

$\dfrac { 1 }{ 2 } \left( KC \right) =\left( 1-\dfrac { 1 }{ K } \right)$

$\left[ \omega \right] =\dfrac { { \epsilon_o }{ AV }^{ 2 } }{ 2d } \left( 1-\dfrac { 1 }{ K } \right)$

In the figure show the potential difference between hollow spheres

$A$ and

$B$ is:-

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zero
0%
$- \dfrac{q}{16 \pi \varepsilon_0 R}$
0%
$\dfrac{q}{16 \pi \varepsilon_0 R}$
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$\dfrac{- q}{24 \pi \varepsilon_0 R}$

Consider a capacitor-charging circuit. Let

$Q_1$ be the charge given to the capacitor in a time interval of

$10ms$ and

$Q_2$ be the charge given in the next time interval of

$10ms$ . Let

$10 \mu C$ charge be deposited in a time interval

$t_1$ and the next

$10 \mu C$ charge is deposited in the next time interval

$t_2$ .

0%
$Q_1 > Q_2, \, t_1 > t_2$
0%
$Q_1 > Q_2, \, t_1 < t_2$
0%
$Q_1 < Q_2, \, t_1 > t_2$
0%
$Q_1 < Q_2, \, t_1 < t_2$

The potential difference between points

$A$ and

$D$ in the given circuit is:-

0%
$\dfrac{2}{3}V$
0%
$\dfrac{8}{9}V$
0%
$\dfrac{4}{3}V$
0%
$2V$

Explanation

Hint: The potential drop on parallel connection is same.

Step 1: Equivalent resistance through ADC,

The equivalent resistance in the series connection is defined as,

$R=R_1+R_2+R_3$

$R=5\Omega+5\Omega+5\Omega$

$R=15\Omega$

Step 2: Finding the current through ADC,

As the potential drop in the parallel connection, therefore the potential drop across

$15\Omega$ is same.

According to Ohm's law,

$V=IR$

$I=\dfrac{V}{R}$

$I=\dfrac{2}{15}$

Step 3: Finding the potential drop.

Using the Ohm's law, the potential drop across AD is,

$V=IR$

$V=\dfrac{2}{15}\times 5$

$V=\dfrac{2}{3}$

Two long straight wires equal cross- sectional radii

$a$ are located parallel to each other in air. The distance between their axes equal

$b$ . Find the mutual capacitance of the wire per unit length under the condition

$b>>a$ .

0%
$C=\dfrac{\pi\varepsilon _0}{ln(b/a)}$
0%
$C=\dfrac{\pi \varepsilon _0}{ln(a/b)}$
0%
$C=\pi/\varepsilon _0In(b/a)$
0%
$C=\pi\varepsilon_0ln(ba)$

what is the series combination of condenses and

$\dfrac { 1 }{ c } =\dfrac { 1 }{ { c }_{ 1 } } +\dfrac { 1 }{ { c }_{ 2 } } +\dfrac { 1 }{ { c }_{ 3 } }$ farad

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True
0%
False

A point charge +Q is located at out side point, at a distance of r from the centre of an uncharged conducting sphere of radius R. Find the electric field and electric potential at the centre of the sphere due to induced charges on the sphere ?

0%
$\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \quad \frac { Q }{ { R }^{ 2 } }$ and Zero
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Zero and $\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \quad \frac { Q }{ { R } }$
0%
$\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \quad \frac { Q }{ { R }^{ 2 } }$ and $\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \quad \frac { Q }{ { R } }$
0%
Zero and Zero

The potential at a point x (measured in

$\mu m$ ) due to some charges situated on the x -axis is given by

$V(x) =\dfrac{20}{(x^{2} -4)}$ volt. The electric field

$E$ at

$x = 4\mu m$ is given by

0%
$\left(\dfrac{10}{9}\right) V\mu m^{-1}$ and in the +ve x direction
0%
$\left(\dfrac{5}{3}\right) V\mu m^{-1}$ and in the -ve x direction
0%
$\left(\dfrac{5}{3}\right) V\mu m^{-1}$ and in the +ve x direction
0%
$\left(\dfrac{10}{9}\right) V\mu m^{-1}$ and in the -ve x direction

A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3d. The slab is equidistant from the plates. The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3 d:

0%
The magnitude of the electric field remains the same
0%
The direction of the electric field remains the same
0%
the electric potential increase continuously.
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The electric potential increase at first, then decreases and again increase

The electric potential in a certain region is expressed by

$V = 6x-8xy^2-8y + 6yz - 4z^2$ volts. The magnitude of the force acting on a charge of

$2C$ sityuated at the origin wil be:-

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$2N$
0%
$6N$
0%
$8N$
0%
$20N$

Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is

$2\times\ 10^5\ V/m$ . When the space is filled with dielectric, the electric field becomes

$1\times\ 10^5\ V/m$ . The dielectric constant of a dielectric material is:

0%
$1/2$
0%
$1$
0%
$2$
0%
$3$

A capacitor 1 mF withstands a maximum voltage of 6KV while another capacitor 2 mF withstands a maximum voltage of 4 KV. If the capacitors are connected in series, the system will withstand a maximum voltage of (MNR)

0%
2 KV
0%
4 KV
0%
6 KV
0%
9 KV

Three point charges of

$1C, 2C$ and

$3C$ are placed at the corners of an equilateral triangle of side

$100\ cm$ . The work done to move these charges to the corners of a similar equilateral triangle of side

$50\ cm$ , will be

0%
$9.9\times 10^{10}J$
0%
$9.9\times 10^{9}J$
0%
$5.2\times 10^{10}J$
0%
$5.9\times 10^{9}J$

In the figure a potential of +1200 V is given to point A and point B is earthed,what is the potential at the point P

0%
100 V
0%
200 V
0%
400 V
0%
600 V

Explanation

Potential of

$=+1200V$

Point

$A$ and point

$B$ is earthed

${ R }_{ 0 }$ at point

$P$ ,

equivalent capacitance,

$\dfrac { \left( 4+2 \right) \times 3 }{ 6+3 } =\dfrac { 6\times 3 }{ 9 } =2\mu F$

Now,

${ V }_{ P }$ at point

$P$

$={ V }_{ P }=1200\times \dfrac { 1 }{ 3 } =400V$

${ V }_{ P }$ is passing through only

$3\mu F$ .

1240526_1212347_ans_59b6fa5fde2643a8aa101ce81a27d4f5.png

A straight conductor of length

$l$ carrying a current t, is bent in the form of a semi-circle, The magnetic field in tesla at the center of the semi-circle is.

0%
$\frac{{{\pi ^2}i}}{l} \times {10^{ - 7}}$
0%
$\frac{{i\pi }}{l} \times {10^{ - 7}}$
0%
$\frac{{i\pi }}{{{l^2}}} \times {10^{ - 7}}$
0%
$\frac{{{\pi ^2}}}{l} \times {10^{ - 7}}$
0%
None of the above.

A parallel plate capacitor has plate area

$A$ and separation

$d$ . It is charged to a potential difference

$V_0$ . The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is:

0%
$\dfrac{3\epsilon_0 AV_0^2}{d}$
0%
$\dfrac{\epsilon_0 AV_0^2}{2d}$
0%
$\dfrac{\epsilon_0 AV_0^2}{3d}$
0%
$\dfrac{\epsilon_0 AV_0^2}{d}$

Explanation

$\textbf{Hint:}$ Use work energy theorem.

$\textbf{Step 1:}$ Work required can be written as,

$W= U_f -U_i$

where,

$U_i=\dfrac{1}{2}CV_0 ^2$

$U_f=\dfrac{1.C.(3V_0)^2}{2.3}$

$U_f=3\times \dfrac{1}{2}CV_0$

$\textbf{Step 2:}$ So work,

$W=\dfrac{\epsilon_0AV_0 ^2}{d}$

A thin metal plate P is inserted half way between the plates of a parallel plate capacitor of capacitance C in such a way that it is parallel to the two plates. The capacitance now becomes

0%
C
0%
C/2
0%
4C
0%
None of these

In given circuit when switch

$S$ has been closed then charge on capacitor

$A$ &

$B$ respectively

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$3\ q, 6\ q$
0%
$6\ q,3\ q$
0%
$4.5\ q, 4.5\ q$
0%
$5,4\ q,4\ q$

Two condemners of capacity

$0.3\mu\ F$ and

$0.6\mu\ F$ respectively are connected in series. The combination is connected across a potential of

$6\ volts$ . The ratio of energies stared by the condensers will be

0%
$\dfrac{1}{2}$
0%
$2$
0%
$\dfrac{1}{4}$
0%
$4$

A parallel plate capacitor consist of two circular plates each of radius 2 cm, separated by a distance of 0.1 mm. If voltage across the plates is varying at the rate of

$5 \times {10^{13}}V{s^{ - 1}}$ , then the value of displacement current is:

0%
$5.50A$
0%
$5.56 \times 10^2 A$
0%
$5.56 \times 10^3 A$
0%
$2.28 \times 10^4 A$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 14 - 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 14 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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