$$\dfrac{2kA\varepsilon_0}{(k+1)d}$$

$$\dfrac{2kA\varepsilon_0}{d}$$

$$\dfrac{(k+1)A\varepsilon_0}{2d}$$

$$\dfrac{2kA\varepsilon_0}{(k^2+1)d}$$

MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 15

A charge of

$10^{-9}C$ is placed on each of the 64 identical drops of radius 2cm.They are then combined to form a bigger drop.Find its potential

0%
$7.2\times10^{3}V$
0%
$7.2\times10^{2}V$
0%
$1.44\times10^{2}V$
0%
$1.44\times10^{3}V$

The total capacitance between point A and B in the arrangement shown below is

0%
$28 \mu F$
0%
$\frac {34}{7} \mu F$
0%
$23 \mu F$
0%
$\frac {34}{3} \mu F$

A charge

$10$ esu is placed at a distance of

$2$ cm, from a charge

$40$ esu and

$4$ cm. from another charge -

$20$ esu. The potential energy of the charge

$10$ esu is :- (In ergs)

0%
$87.5$
0%
$112.5$
0%
$250$
0%
$zero$

In a region of uniform field

$,$ as an electron travels from

$A$ to

$B,$ it slows from

${u_A} = 6.1 \times {10^6}\,m/s$ to

${u_B} = 4.5 \times {10^6}\,m/s.$ What is its potential change

$\Delta V = {V_B} - {V_A}$ in volts?

0%
$18$
0%
$-18$
0%
$+48$
0%
$-48$

The polarising angle of a glass slab for green light is

$54.75^0$ . what is the angle of minimum deviation of a ray of green light for an equilateral prism made of the same material as that of the glass slab?

$( given : \tan 54.75^0 =1.414 )$

0%
$60^0$
0%
$54.75^0$
0%
$45^0$
0%
$30^0$

"The equipotential surface will never intersect each other". This statment is

0%
True
0%
False
0%
May be true
0%
Can't be determined

The circuit shown is in state. Electrostatic potential energy stored by the circuit is

0%
${ CV }^{ 2 }$
0%
$3 { CV }^{ 2 }$
0%
$5 { CV }^{ 2 }$
0%
$7 { CV }^{ 2 }$

If electric potential of a body is zero, than

0%
Charge on this must be zero
0%
Charge on this may be positive
0%
Charge on this may be negative
0%
Both $(B)$ & $(C)$

In the figure shown

$a+3 \mu c$ charge and

$a-2 \mu c$ charge are placed on the x-axis.in which of the regions on the x-axis is a point where the electrical potential is zero.

0%
Region II only
0%
Region III only
0%
Region II and III
0%
Region I, II and III

In the figure shown

$a+3 \mu c$ charge and

$a-2 \mu c$ charge are placed on the x-axis.in which of the regions on the x-axis is a point where the electrical potential is zero.

0%
Region II only
0%
Region III only
0%
Region II and III
0%
Region I, II and III

An air capacitor is connected to a battery of emf V. A dielectric medium of dielectric constant K is introduced between the plates of this capacitor, with the battery still connected. The final charge on the capacitor plates relative to the previous charge

0%
is less
0%
is more
0%
is the same
0%
may be more or less depending upon the dielectric constant of the medium

Explanation

REF.Image

Initially

$Q = CV$

$= \frac{A\varepsilon _{0}}{d}V$

$(k=1)$

But when dielectric

is inserted

$Q^{1}=K\frac{A\varepsilon _{0}}{d}V$

$k > 1$

$Q^{1}> Q$

(B) option is correct.

1194746_1275788_ans_ebb844186fac497bb82103c77032f924.jpeg

The ratio of the capacitance

$\dfrac{C_{1}}{C_{2}}$ will be

0%
$2/3$
0%
$4/3$
0%
$3/4$
0%
$3/2$

The potential in certain region is given as

$V = 2x^2$ , then the charge density of that region is

0%
$-\dfrac{4x}{\varepsilon _0}$
0%
$-\dfrac{4}{\varepsilon _0}$
0%
$-4 \varepsilon_0$
0%
$-2 \varepsilon_0$

Explanation

By Gauss's law,

$\nabla^2V=-\frac{\rho}{\varepsilon_0}$

So,

$\frac{\partial^2V}{\partial x^2}=-\frac{\rho}{\varepsilon_0} ...(1)$

Give,

$V=2x^2$

$\frac{\partial V}{\partial x}=4x$

$\frac{\partial^2V}{\partial x^2}=4$

Now from (1),

$\rho=-4\varepsilon_0$

A thin diamgnetic ros is placed vertically between the poles of an elctromagnet. when the current in the electromagnet is switched on, then the dimagnetic rod is pushed up, out of the horizontal magnetic field. hence the rod gains gravitational potential energy. the work required to do this comes from

0%
The magnetic field
0%
The lattice structure of the material of the rod
0%
The current source
0%
The induced electric field due to the changing magnetic field

The equivalent capacitance in the circuit between

$A$ and

$B$ will be

0%
$\dfrac{1}{3} \,\mu F$
0%
$3 \,\mu F$
0%
$2 \,\mu F$
0%
$1\,\mu F$

Two equal point charges A and B are R distance apart. A third point charge placed on the perpendicular bisector at a distance d from the centre will experience maximum electrostatic force when

0%
$d=\frac{R}{2 \sqrt 2}$
0%
$d=\frac{R}{ \sqrt 2}$
0%
$d=R \sqrt 2$
0%
$d=2 \sqrt 2 R$

For section

$AB$ of a circuit shown in Figure.

$C_1=1\mu F$ ,

$C_2=2\mu F$ ,

$E=10V$ , and the potential difference

$V_A-V_B=-10V$ . Charge on capacitor

$C_1$ is

0%
$0\mu C$
0%
$20/3\mu C$
0%
$40/3\mu C$
0%
$None$

The equivalent capacitance between the points A and B of a combination shown in the figure is

0%
C
0%
2C
0%
C/2
0%
3C

the plates of a parallel plate capacitor are separated by a distance d with air as the medium between the plates.in order to increase the capacitance by 66% a dielectric slab of dielectric constant 5 is introduced between the plates what is the thickness of the sielectric slab?

0%
$\frac {d}{4}$
0%
$\frac {d}{2}$
0%
$\frac {5d}{8}$
0%
$d$

A very narrow tunnel is formed along the diameter of a spherical uniformly charge non-conducting body. Its radius is

$R$ and total charge is

$Q$ . If an electron is released from rest from one end of the tunnel then kinetic energy of electron at centre of sphere is

$\left( k=\dfrac { 1 }{ 4\pi \varepsilon _{ 0 } } \right)$

0%
$\dfrac { kQe }{ 2R }$
0%
$\dfrac { kQe }{ \sqrt { 2R } }$
0%
$\dfrac { kQe }{ 4R }$
0%
$\dfrac { kQe }{ \sqrt { 3R } }$

An electric field of

$200V{m^{ - 1}}$ exists in the region between the plates of a parallel plate capacitor of plate separation

$5cm$ . The potential difference between the plates when a slab of dielectric constant

$4$ and thickness

$1cm$ is inserted between the plates is:

0%
$7.5V$
0%
$8.5V$
0%
$9.0V$
0%
$10V$

An inductor

$\left( L=\frac { 1 }{ 100\pi } H \right)$ ,a capacitor

$\left( C=\frac { 1 }{ 500\pi } F \right)$ and a resistance

$(3\Omega )$ is connected in series with an AC voltage source as shown in the figure. The voltage of the AC source is given as

$V=10cos(100\pi t)$ volt. What will be the potential difference between A and B ?

0%
$8cos(100\pi t-{ 127 }^{ 0 })volt$
0%
$8cos(100\pi t-{ 53 }^{ 0 })volt$
0%
$8cos(100\pi t-{ 37 }^{ 0 })volt$
0%
$8cos(100\pi t+{ 37 }^{ 0 })volt$

Electric potential

$(\phi )$ of a quadrupole varies with distance 'r' on its axis as :-

0%
$\phi :{ r }^{ -1 }$
0%
$\phi :{ r }^{ -2 }$
0%
$\phi :{ r }^{ -3 }$
0%
$\phi :{ r }^{ { 3 }/{ 2 } }$

The correct graph between the gravitational potential

$(v_{g})$ due to a hollow sphere and distance from centre will be

The electric potential at point

$P$ is.

0%
$8\ V$
0%
$6\ V$
0%
$10\ V$
0%
$4\ V$

Explanation

The electric field between charged parallel plates is uniform, which means the potential charges uniformly with distance. For a change of

$8V$ ones

$4cm$ means the change of potential with position (and the electric field strength) is

$2V/cm$ . Which gives the potential

$1cm$ away from

$2V$ plate as

$4V$ .

1241888_1325952_ans_5dbf848260714e4795ec709910117cfe.png

Find a capacitance

0%
$\dfrac{2kA\varepsilon_0}{d}$
0%
$\dfrac{2kA\varepsilon_0}{(k+1)d}$
0%
$\dfrac{(k+1)A\varepsilon_0}{2d}$
0%
$\dfrac{2kA\varepsilon_0}{(k^2+1)d}$

Potential is varying with x and y as v=2

$\left( { x }^{ 2 }{ -y }^{ 2 } \right) .$ The corresponding field pattern is:

Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of

$\varepsilon _r=4$ .

0%
Calculate capacitance of each capacitor if equivalent capacitance of the combination is $4 \mu F$ .
0%
Calculate the potential difference between the plates of X and Y.
0%
Estimate the ratio of electrostatic energy stored in X and Y.
0%
None of these

Two concentric conducting shells have radii r and R ( r < R) respectively. The inner sphere has charge q and outer sphere is uncharged . The potential difference between the two spheres is V. If the outer sphere is given a charge 5q , then the potential difference between the spheres will be

0%
5 V
0%
6V
0%
2V
0%
V

Condenser A has a capacity of 15

$\mu F$ when it is filled with a medium of dielectric constantAnother condenser B has a capacity 1

$\mu F$ with air between the plates. Both are charged separately by a battery of 100V. After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is

0%
400 V
0%
800 V
0%
1200 V
0%
1600 V

In a parallel-plate capacitor with plate area

$A$ and charge

$Q$ the force on one plate because of the charge on the other is equal to.

0%
$\dfrac{Q^2}{\varepsilon_0 A^2}$
0%
$\dfrac{Q^2}{2\varepsilon_0 A^2}$
0%
$\dfrac{Q^2}{\varepsilon_0 A}$
0%
$\dfrac{Q^2}{2\varepsilon_0 A}$

The magnitude of the electric field at point

$P$ is:

0%
$\text{800 V/m}$
0%
$\text{600 V/m}$
0%
$\text{400 V/m}$
0%
$\text{200 V/m}$

A,B,C,D are four large metallic plates carrying charges

$+4 \mu C, -3 \mu C, 0, \mu C and +2 \mu C$ placed close and parallel to each other as shown.The charge (in equilibrium) retained upon left side of left most plate is

0%
$3.00 \mu C$
0%
$0.75 \mu C$
0%
$1.5 \mu C$
0%
$-1.5 \mu C$

A wire, bent into a circle, carries current in an anticlockwise direction. What polarity does this face of the coil exhibit?

0%
North
0%
East
0%
West
0%
South

The figure shows a capacitor having three layers of equal thickness and same area is that of plate . Layer first is vaccum , layer second is conductor and layer third is dielectric of dielectric constant

$K$ . The ratio of energy stored in region three to the total energy stored in capacitor is:

0%
$\dfrac { 1 } { K + 1 }$
0%
$\dfrac { 3 } { K + 1 }$
0%
$\dfrac { 4 } { K + 3 }$
0%
$\dfrac { 4 } { 3K + 1 }$

ABCD is a square where each each side is a uniform wire of resistance

$1 \Omega$ . A point E lies on CD such that is a uniform wire of resistance

$1 \Omega$ is connected across AE and constant potential difference is applied across A and C , then B and E are equipotential.

0%
$\frac {CE}{ED}=1$
0%
$\frac {CE}{ED}=\frac {1}{\sqrt2}$
0%
$\frac {CE}{ED}=\frac {1}{2}$
0%
$\frac {CE}{ED}=\dfrac{\sqrt 2}{1}$

Explanation

Wire of resistance

$=1\Omega$

Keeping in view that,

$ABCD$ is a square where each side is uniform wire of resistance

$1\Omega$ . If a uniform wire of resistance

$1\Omega$ is connected across

$AE$ and a potential difference is applied across

$A$ and

$C$ , the points

$B$ and

$E$ will be equipotential; a point

$E$ on

$CD$ is

$\dfrac { CE }{ ED } =\dfrac { \sqrt { 2 } }{ 1 }$

$\therefore$

$CE:ED=\sqrt { 2 } :1$

1244036_1388831_ans_6ac66432af014c9bb4cd279e335b9dda.png

In the circuit shown, the potential difference between A and B is:

0%
6V
0%
1V
0%
2V
0%
3V

Charges

$q_1, q_2\ and\ q_3$ are placed on capacitor of capacitance

$C_1, C_2\ and\ C_3$ respectively arranged in series as shown, switch S is then closed.

$Given\ q_1 = 30 C, q_2 = 20 C, q_3 = 10 C; C_1 = 10 F; C_2 = 20F; C_1 = 30 F\ and\ 12 volts$
The final chage on capacitor

$C_1 is-$

0%
$q_1 = \frac {690}{11} \mu C$
0%
$q_1 = \frac {790}{11} \mu C$
0%
$q_1 = \frac {5000}{11} \mu C$
0%
$q_1 = \frac {550}{11} \mu C$

In the given circuit, find the potential difference across the 6

$\mu \mathrm { F }$ capacitor in steady state.

0%
$4 V$
0%
$2 V$
0%
$6 V$
0%
$8 V$

Explanation

$\begin{array}{l} At\, \, steady\, \, state\, \, all\, \, the\, \, capacitor\, \, acts\, \, as\, \, the\, \, open\, \, circuit. \\ find\, \, the\, \, difference\, \, { V_{ CD } } \\ resis\tan ce\, \, across\, \, CD=\left( { 1.5+1.5 } \right) \parallel \left( { 1.5+1.5 } \right) +1.5=3\, \Omega \\ { V_{ CD } }=8\left( { \frac { 3 }{ 4 } } \right) =6\, V \\ potential\, \, difference\, \, across\, \, 6\mu F=6\left( { \frac { 3 }{ 9 } } \right) =2V \end{array}$

Hence,

option

$(B)$ is correct answer.

A non-conducting sphere has a total charge Q uniformly distributed throughout its volume. the centre of the sphere is at origin and its radius is R. let

$U_1$ be the electrostatic potential energy in the region inside the sphere and

$U_2$ be the electrostatic potential energy in another imaginary spherical shell, having inner radius R and outer radius infinity, centred at origin. select the correct alternative(s)

0%
$U_1 = \dfrac {Q^2}{8 \pi \varepsilon_0 R}$
0%
$U_2 = \dfrac {Q^2}{8 \pi \varepsilon_0 R}$
0%
$U_1 + U_2 = \dfrac {3}{5} \times \dfrac {Q^2}{4 \pi \varepsilon_0 R}$
0%
$U_1 = \dfrac {3Q^2}{20 \pi \varepsilon_0 R }$

Explanation

Non-conducting sphere has a total charge

$=Q$

the centre of the sphere is at origin and its radius

$=R$

${ U }_{ 1 }=$ Electrostatic potential energy in the region inside the sphere

${ U }_{ 2 }=$ the electrostatic potential energy in another spherical shell.

having inner radius

$=R$

outer radius infinity

${ U }_{ 1 }=\dfrac { { Q }^{ 2 } }{ \delta \pi { \epsilon }_{ 0 }R }$ because, the potential will be

${ U }_{ 1 }=\dfrac { 1 }{ 2 } Q\times U$ (This is the working formula of potential energy)

${ U }_{ 1 }=\dfrac { 1 }{ 2 } \times Q\times \dfrac { Q }{ 4\pi { \epsilon }_{ 0 }R }$

or,

${ U }_{ 1 }=\dfrac { 1 }{ 2 } \times \dfrac { { Q }^{ 2 } }{ 4\pi { \epsilon }_{ 0 }R } =\dfrac { { Q }^{ 2 } }{ 8\pi { \epsilon }_{ 0 }R }$

Since,

${ U }_{ 2 }$ be the electrostatic potential energy then,

${ U }_{ 2 }=0$ .

$5\mu F$ is charged to

$12\ volt$ . The positive plate of The positive plate of the capacitor is connected to the negative terminal of a

$12\ V$ battery. Then heat developed in the connecting wire will be:

0%
$2.44\ mJ$
0%
$1.44\ mJ$
0%
$0.44\ mJ$
0%
$3.8\ mJ$

$1\ MeV$ equals

0%
$1.6\times 10^ {-19}\ J$
0%
$1.6\times 10^ {-16}\ J$
0%
$1.6\times 10^ {-13}\ J$
0%
$1.6\times 10^ {13}\ J$

A charge of

$10 \mu C$ is placed at the origin of

$x - y$ coordinate system. The potential difference between two points

$( 0 , \text { a) and } ( a , 0 )$ in volt will be

0%
$\dfrac { 9 \times 10 ^ { 4 } } { a }$
0%
$\dfrac { 9 \times 10 ^ { 4 } } { a \sqrt { 2 } }$
0%
$\dfrac { 9 \times 10 ^ { 4 } } { 2 a }$
0%
$Zero$

Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

0%
$\dfrac{C}{3}, \dfrac{V}{3}$
0%
$3C, \dfrac{V}{3}$
0%
$\dfrac{C}{3}, 3V$
0%
$3C, 3V$

A particle having charge that on an electron and mass

$1.6 \times 10^{-30}$ kg is projected with an initial speed 'u' to the horizontal from the lower plate of a parallel plate capacitor as shown. The plates are sufficiently long and have separation 2cm. Then the maximum value of velocity of particle not to hit the upper plate. (

$E = 10^3 V/m$ upwards)

0%
$2 \sqrt{2} \times 10^6m/s$
0%
$4 \times 10^6 m/s$
0%
$6 \times 10^6 m/s$
0%
$3 \times 10^6 m/s$

Two conductors, each of capacitance 1F are charged to potential of 10 V and 6 V respectively. They are then joined together. Their common potential will be

0%
16 V
0%
8 V
0%
4 V
0%
1 V

A charged cloud system produces an electric field in the air near the earth's surface. A particle of charge

$-2 \times 10^{-9} \mathrm{C}$ is acted on by a downward electrostatic force of

$3 \times 10^{-6} \mathrm{N}$ when placed in this field. The gravitational and electrostatic force, respectively, exerted on a proton placed in this field are

0%
$1.64 \times 10^{-26} \mathrm{N}, 2.4 \times 10^{-16} \mathrm{N}$
0%
$1.64 \times 10^{-26} \mathrm{N}, 1.5 \times 10^{3} \mathrm{N}$
0%
$1.56 \times 10^{-18} \mathrm{N}, 2.4 \times 10^{-16} \mathrm{N}$
0%
$1.5 \times 10^{3} N, 2.4 \times 10^{-16} \mathrm{N}$

A resistor

$^{\prime} R^{\prime}$ and

$2 \mu F$ capacitor in series is connected through a switch to

$200 \mathrm{V}$ direct supply. Across the capacitor is a neon bulb that lights up at

$120 \mathrm{V}$ Calculate the value of

$R$ to make the bulb light up

$5 s$ after the switch has been closed.

$\left(\log _{10} 2.5=0.4\right)$

0%
$2.7 \quad 10^{6} \Omega$
0%
$3.3 \quad 10^{7} \Omega$
0%
$1.3 \quad 10^{4} \Omega$
0%
$1.7 \quad 10^{5} \Omega$

In the figure, a proton moves a distance d in a uniform

$\vec{E}$ as shown in the figure. If W is work done by the electric field and

$\Delta U$ is change in the electric potential energy of the proton then.

0%
W < 0 and $\Delta U$ > 0
0%
W < 0 and $\Delta U$ < 0
0%
W > 0 and $\Delta U$ > 0
0%
W < 0 and $\Delta U$ = 0

Two capacitors of capacitances

$4\mu F$ and

$6\mu F$ are connected across a 120 V battery in series with each other. What is the potential difference across the

$4\mu F$ capacitor?

0%
40V
0%
48V
0%
60V
0%
72V

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 15 - 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 15 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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