MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 13

A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one fourth of the capacitor as shown in the figure. The new capacitance will be :

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$\dfrac{(K + 3) C}{4}$
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$\dfrac{(K + 2) C}{4}$
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$\dfrac{(K + 1) C}{4}$
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$\dfrac{KC}{4}$

Explanation

Capacitance

$=\dfrac{\epsilon A}{d}$

Initially for air:

$C=\dfrac{\epsilon_o A}{d}$

Finally for air:

$C=\dfrac{\epsilon_o \dfrac{3A}{4}}{d}=\dfrac{3C}{4}$

For dielectric:

$C_k=\dfrac{K\epsilon_o \dfrac{A}{4}}{d}=\dfrac{KA}{4}$

We can consider them as two capacitors in series which gives,

$C_{net}=\dfrac{3C}{4}+\dfrac{KC}{4}=\dfrac{(K+3)C}{4}$

Equipotential surfaces associated with an electric field which is increasing in magnitude along the x-direction are:

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Planes parallel to yz-plane
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Planes parallel to xy-plane
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Planes parallel to xz-plane
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Coaxial cylinders of increasing radii around

the x-axis

Explanation

We know that

$\vec{E}=-\vec{\nabla}\phi$

Thus the equipotential surface is always perpendicular to the direction of electric field. As the field is along x-direction, equipotential surface must be parallel to yz-plane.

Assertion: Charges are given to plates of two plane parallel plate capacitors

$\displaystyle { C }_{ 1 }$ and

$\displaystyle { C }_{ 2 }$ (such that

$\displaystyle { C }_{ 2 }=2{ C }_{ 1 }$ )as shown in figure. Then the key

$K$ is pressed to complete the circuit. Finally the net charge on upper plate and net charge the circuit. Finally the net charge on upper plate and net charge on lower plate of capacitor

$C$ is positive.
Reason : In a parallel plate capacitor both plates always carry equal and opposite charge.

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If both Assertion arid Reason are correct and Reason is the correct explanation of Assertion.
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If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
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If Assertion is correct but Reason is incorrect.
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If Assertion is incorrect but Reason is correct.

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant

$K$ . The remaining half contains air as shown in the figure. The capacitor is now given a charge

$Q$ . Then :

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Electric field in the dielectric-filled region is higher than that in the air-filled region
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On the two halves of the bottom plate the charge densities are unequal
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Charge on the half of the top plate above the air-filled part is $\dfrac { Q }{ K+1 }$
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Capacitance of the capacitor shown above is $\left( 1+K \right) \dfrac { { C }_{ 0 } }{ 2 }$ , where ${ C }_{ 0 }$ is the capacitance of the same capacitor with the dielectric removed.

Explanation

We know that

${ C }_{ 1 }=\dfrac { K{ \varepsilon }_{ 0 }A }{ 2d } , { C }_{ 2 }=\dfrac { { \varepsilon }_{ 0 }A }{ 2d }$ Where

$C_1$ is capacitance with dielectric and

$C_2$ without dielectric.
and

${ C }_{ eq }=\dfrac { \varepsilon \cdot A }{ 2d } \left( K+1 \right)$

${ C }_{ eq }=\dfrac { { C }_{ 0 } }{ 2 } \left( K+1 \right)$

$\dfrac { { Q }_{ 1 } }{ { Q }_{ 2 } } =\dfrac { { C }_{ 1 } }{ { C }_{ 2 } } =\dfrac { K }{ 1 }$

$\Rightarrow \dfrac { { \sigma }_{ 1 } }{ { \sigma }_{ 2 } } =\dfrac { K }{ 1 }$

Where Q1 and Q2 are the charges stored in the capacitors.

${ Q }_{ 1 }=\dfrac { KQ }{ K+1 }$ and

${ Q }_{ 2 }=\dfrac { Q }{ K+1 }$
So,

$E=\dfrac { \sigma }{ { \varepsilon }_{ 0 }K } \Rightarrow \dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { { \sigma }_{ 1 } }{ { \sigma }_{ 2 } } \times \dfrac { { K }_{ 1 } }{ 1 }$

$=\dfrac { { Q }_{ 1 } }{ { Q }_{ 2 } } \times \dfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\dfrac { K }{ 1 } \times \dfrac { 1 }{ K }$

$= 1 : 1$

Two charges of equal magnitude

$q$ are placed in air at a distance

$2a$ apart and third charge

$-2q$ is placed at mid-point. The potential energy of the system is (

${\varepsilon}_{0} =$ permittivity of free space) :

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$-\dfrac{{q}^{2}}{8\pi {\varepsilon}_{0} a}$
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$-\dfrac{3{q}^{2}}{8\pi {\varepsilon}_{0} a}$
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$-\dfrac{5{q}^{2}}{8\pi {\varepsilon}_{0} a}$
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$-\dfrac{7{q}^{2}}{8\pi {\varepsilon}_{0} a}$

Explanation

Potential energy of the system

$U=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } +\dfrac { { q }_{ 1 }{ q }_{ 3 } }{ { r }_{ 13 } } +\dfrac { { q }_{ 2 }{ q }_{ 3 } }{ { r }_{ 23 } } \right]$

$=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { q\left( -2q \right) }{ a } +\dfrac { q\left( -2q \right) }{ a } +\dfrac { qq }{ 2a } \right]$

$=\dfrac { 1 }{ 4{ \pi \varepsilon }_{ 0 } } \left[ -\dfrac { 2{ q }^{ 2 } }{ a } -\dfrac { 2{ q }^{ 2 } }{ a } +\dfrac { { q }^{ 2 } }{ 2a } \right]$

$=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { -4{ q }^{ 2 } }{ a } +\dfrac { { q }^{ 2 } }{ 2a } \right] =\dfrac { -7{ q }^{ 2 } }{ 8\pi { \varepsilon }_{ 0 }a }$

A parallel plate capacitor has a capacity

$80\times 10^{-6}\ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant

$20$ . The capacitor is now connected to a battery of

$30\ V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is :

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$45.6\times 10^{-3}\ C$
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$25.3\times 10^{-3}\ C$
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$120\times 10^{-3}\ C$
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$12\times 10^{-3}\ C$

Explanation

Charge passing through the wire,

$\Delta q=\Delta CV$

$=(C'-C)V\\=(k-1)C V\\=(20-1)(80\times 10^{-6})(30)\\=45.6\times 10^{-3}$

So, charge passing through the wire

$=45.6 \times 10^{-3}\ C$

Consider a parallel plate capacitator of capacity 10

$\displaystyle \mu F$ filled with air. When the gap between the plates is filled partly with a dielectric of dielectric constant

$4$ , as shown in figure, the new capacity of the capacitator is (

$A$ is the area of plates):

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$20 \displaystyle \mu F$
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$40\displaystyle \mu F$
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$2.5\displaystyle \mu F$
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$25\displaystyle \mu F$

Explanation

Such a configuration can be thought of as a parallel combination of two capacitors- one with area A/2 and dielectric and another with same area and without dielectric.

${ C }_{ 1 }=\dfrac { { \epsilon }_{ 0 }A }{ 2d }$ ; (for air in between)

In second case

$k = 4$

${ C }_{ 2 }=\dfrac { 4{ \epsilon }_{ 0 }A }{ 2d } =\dfrac { 2{ \epsilon }_{ 0 }A }{ d }$

Hence, effective capacitance is

$C={ C }_{ 1 }+{ C }_{ 2 }=2.5\dfrac { { \epsilon }_{ 0 }A }{ d } =2.5{ C }_{ 0 }=25\ \mu F$

The plate separation in a parallel plate condenser is

$d$ and plate area is

$A.$ If it is charged to

$V$ volts and a battery is disconnected then work done in increasing the plate separation to

$2d$ will be

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$\cfrac{3}{2}\cfrac{\epsilon_oAV^2}{d}$
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$\cfrac{\varepsilon _oAV^2}{d}$
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$\cfrac{2\varepsilon _oAV^2}{d}$
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$\cfrac{\varepsilon _oAV^2}{2d}$

Explanation

Workdone

$=\cfrac{1}{2}C_2V^2-\cfrac{1}{2}C_1V^2$
As

$d$ increases, capacitance change

$W=\cfrac{1}{2}(C_2-C_1)V^2$

$C_2=\cfrac{\epsilon_oA}{2d},\quad C_1=\cfrac{\epsilon_oA}{d}$

$W=\cfrac{1}{2}\epsilon_oA\left(\cfrac{1}{2d}-\cfrac{1}{d}\right)V^2$

$W=-\cfrac{\epsilon_oA}{2d}V^2$
Negative sign shows external energy given from outside to do work.

Two identical condensers M and N are connected in series with a battery. The space between the plates of M is completely filled with a dielectric medium of dielectric constant

$8$ and a copper plate of thickness

$\displaystyle \frac { d }{ 2 }$ is introduced between the plates of N (

$d$ is the distance between the plates). Then potential differences across M and N are, respectively, in the ratio:

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$1:4$
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$4:1$
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$3:8$
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$1:6$

Explanation

Let the capacitance of the condenser

$M$ be

$C$ and

$N$ be

$C'$ .

The new capacitance of the condenser

$M$ will be,

$C_M=\varepsilon C_M=8C$

When a copper plate of thickness

$d/2$ is inserted in between the capacitor N, effective distance between plates reduces to

$d-d/2=d/2$ .

Since,

$C\propto \dfrac{1}{d}, \quad C'=2C$

Since, the two condensers are connected in series, charge one each capacitor is same.

$\Rightarrow V_M:V_N=\dfrac{1}{C_M}:\dfrac{1}{C_N}=\dfrac{1}{8C}:\dfrac{1}{2C}$

$\Rightarrow V_M:V_N=1:4$

A very thin metal sheet is inserted halfway between the parallel plates of an air-gap capacitor. The sheet is thin compared to the distance between the plates, and it does not touch either plate when fully inserted. The system had capacitance,

$C$ , before the plate is inserted.
What is the equivalent capacitance of the system after the sheet is fully inserted?

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$\cfrac{1}{4}C$
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$\cfrac{1}{2}C$
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$C$
0%
$2C$
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$4C$

Explanation

Initially (before metal sheet inserted) the capacitance of a parallel plate capacitor is

$C=\dfrac{A\epsilon_0}{d}$ where A be the area of plates and d be the separation between parallel plates.
When a metal sheet inserted fully halfway between the parallel plates, the capacitance will be divided into two capacitors

$C_1, C_2$ and they are in series.
Thus,

$C_1=\dfrac{A\epsilon_0}{(d/2)}=2C$ and

$C_2=\dfrac{A\epsilon_0}{(d/2)}=2C$
The equivalent capacitance ,

$C_{12}=\dfrac{C_1C_2}{C_1+C_2}=\dfrac{2C\times 2C}{2C+2C}=C$

Two metal plates each of area

$A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is

$d$ . A metal plate of thickness

$\dfrac{d}{2}$ and of same area

$A$ is inserted between the plates to form two capacitors of capacitances

$C_1$ and

$C_2$ as shown in the figure. The effective capacitance of the two capacitors is C' and the capacitance of the capacitor initially is C, then

$\frac{C'}{C}$ is :

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$4$
0%
$2$
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$6$
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$1$

Explanation

Let the gap for capacitor 1 be x and that for capacitor 2 be y. Then,

${ C }_{ 1 }=\dfrac { { \epsilon }_{ 0 }A }{ x }$ and

${ C }_{ 2 }=\dfrac { { \epsilon }_{ 0 }A }{ y }$

$x+y=d/2$

Since the two are in series,

$C'=\dfrac { { C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } ={ \epsilon }_{ 0 }A\dfrac { \dfrac { 1 }{ xy } }{ \dfrac { 1 }{ x } +\dfrac { 1 }{ y } }$

$=\dfrac { { \epsilon }_{ 0 }A }{ x+y } =\dfrac { 2{ \epsilon }_{ 0 }A }{ d } =2C$

Hence ,

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

Three charges

$Q, + q$ and

$+q$ are placed at the vertices of a right angle triangle (isosceles triangle) as shown. If the net electrostatic potential energy of the configuration is zero, value of

$Q$ is:

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$\displaystyle \frac { +q }{ 2+\sqrt { 2 } }$
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$\displaystyle \frac { -q }{ 2+\sqrt { 2 } }$
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$\displaystyle \frac { +2q }{ 2+\sqrt { 2 } }$
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$\displaystyle \frac { -2q }{ 2+\sqrt { 2 } }$

Explanation

We know that the electrostatic potential energy of a two charge system separated by a distance r is given by,

$E=\dfrac { k{ q }_{ 1 }{ q }_{ 2 } }{ r }$

Hence, for the configuration in figure to have zero energy,

$\dfrac { k{ q }^{ 2 } }{ a } +\dfrac { kQq }{ a } +\dfrac { kQq }{ \sqrt { 2 } a } =0$

Hence,

$Q+\dfrac { Q }{ \sqrt { 2 } } =-q$

$Q=\dfrac { -\sqrt { 2 } q }{ \sqrt { 2 } +1 } =\dfrac { -2q }{ 2+\sqrt { 2 } }$

Within a spherical charge distribution of charge density

$\rho(r)$ , N equipotential surfaces of potential

$V_0, V_0 + \Delta V, V_0 + 2\Delta V, ..... ,V_0 + N\Delta V\,\, (\Delta V > 0)$ , are drawn and have increasing radii

$r_0, r_1, r_2,.....r_N$ , respectively. If the difference in the radii of the surfaces is constant for all values of

$V_0$ and

$\Delta V$ then :

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$\rho (r) =$ constant
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$\rho(r)\propto \dfrac{1}{r^2}$
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$\rho(r) \propto \dfrac{1}{r}$
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$\rho(r)\propto 1$

Explanation

Considering a spherical gaussian surface and applying Gauss law,

$E(4\pi { r }^{ 2 })=\dfrac {\displaystyle \int _{ 0 }^{ r }{ \rho (4\pi { r }^{ 2 })dr } }{ { \epsilon }_{ 0 } }$

$E=\cfrac {\displaystyle \int _{ 0 }^{ r }{ \rho { r }^{ 2 }dr } }{ { \epsilon }_{ 0 }{ r }^{ 2 } }$

It is given that

$\dfrac{dV}{dr}$ is constant.

But

$E=-\dfrac{dV}{dr}$

$\implies E$ is constant

$\implies \displaystyle\int \rho r^2 dr \propto r^2$

$\implies \rho r^2 \propto r$

$\implies \rho \propto \dfrac{1}{r}$

The potential difference between two points A and B is

$10V$ . Point A is at higher potential. If a negative charge, q = 2 C, is moved from point A to point B, then the potential energy of this charge will:

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decrease by 20 J
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decrease by 5 J
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increase by 5 J
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increase by 20 J
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increase by 100 J

Explanation

The change in potential energy,

$\Delta U=q\Delta V$

As the charge

$q=-2C$ is moved from A to B so potential will decrease 10 V i.e

$\Delta V=-10 V$

Thus,

$\Delta U=(-2)(-10) =+20 J$

The electric field at any point

$(x,y,z)$ is

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$(8x-3)\hat i$
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$-(8x-3)\hat i$
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$-8x\hat i$
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$8x\hat i$

Explanation

$V=4x^2-3x$

$\vec{E}=-\dfrac{\partial V}{\partial x}\hat{i}-\dfrac{\partial V}{\partial y}\hat{j}-\dfrac{\partial V}{\partial z}\hat{k}$

Here,

$\dfrac{\partial V}{\partial x}=8x-3$

And,

$\dfrac{\partial V}{\partial y}=\dfrac{\partial V}{\partial z}=0$

Hence,

$\vec{E}=-(8x-3)\hat{i}$

Answer-(B)

A voltmeter reads

$4\ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between plates for the same configuration, voltmeter reads

$2\ V$ . What is the dielectric constant of the material?

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$0.5$
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$2$
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$8$
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$10$

Explanation

For a capacitor with dielectric, capacitance is given by:

$C= \cfrac{K\varepsilon_o A}{d}$

From law of conservation of charge, initial and final charge on the capacitor is same.

$Q_i = Q_f$

From definition of capacitance,

$C_iV_i = C_fV_f$

$\cfrac{\varepsilon_o A}{d}V_i = \cfrac{K\varepsilon_o A}{d} V_f$

$4 = 2K$

$K=2$

Two parallel plate air capacitors each of capacitance

$C$ were connected in series to a battery with e.m.f.

$E$ . Then one of the capacitors was filled up with a uniform dielectric with relative permittivity

$k$ . What amount of charge flows through the battery?

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$\Delta q=\dfrac{1}{2}CE\dfrac{k+1}{k-1}$
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$\Delta q=\dfrac{1}{2}CE\dfrac{k-1}{k+1}$
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$\Delta q=\dfrac{1}{2}CE\dfrac{2k-1}{k+1}$
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None of these

The electric potential

$V$ is given as a function of distance

$x$ by

$V = 5x^2 +10x - 9$ . The value of electric field at

$x=1$ is
(All quantities are in SI units.)

0%
$-20$
0%
$6$
0%
$11$
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$-23$

Explanation

$V=5x^2+10x-9$

$\vec{E}=-\dfrac{\partial V}{\partial x}\hat{i}-\dfrac{\partial V}{\partial y}\hat {j}-\dfrac{\partial V}{\partial z}\hat{k}$

Now,

$\dfrac{\partial V}{\partial x}=10x+10$

and

$\dfrac{\partial V}{\partial y}=\dfrac{\partial V}{\partial z}=0$

Hence,

$\vec{E}=-(10x+10)\hat{i}$

For

$x=1$ ,

$\vec{E}=-20\hat{i}$

Answer-(A)

A particle of mass

$10^{-3}kg$ and charge

$5\mu C$ is thrown at a speed

$20\ m\ s^{-1}$ against a uniform electric field of strength

$2\times 10^{5}N\ C^{-1}$ . How much distance will it travel before coming to rest momentarily?

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$0.1\ m$
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$0.3\ m$
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$0.05\ m$
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$0.2\ m$

Explanation

Force on particle

$=F=qE$ in opposite direction of motion

And,

$F=ma=qE$

$\implies a=\dfrac{qE}{m}=\dfrac{5\times 10^{-6}\times 2\times 10^5}{10^{-3}}$

$\implies a=10^3m/s^2$ and this acceleration is negative since particle is thrown against force.

And final velocity is

$v=0$

Using

$v^2-u^2=2as$

$\implies 0-20^2=-2\times 1000\times s$

$\implies s=0.2m$

Answer-(D)

Electric potential due to a point charge of

$\dfrac{1}{9} pC$ at a distance

$1k m$ from its centre is:

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$0.1V$
0%
$10^{-6}V$
0%
$10^6V$
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$0V$

Explanation

Electric potential is given by

$V=\dfrac{q}{4\pi\epsilon_o r}=\dfrac{kq}{r}$ where

$k=\dfrac{1}{4\pi\epsilon_o}=9\times 10^9$

Hence,

$v=\dfrac{9\times 10^9\times \dfrac{1}{9}\times 10^{-12}}{1000}$

$\implies V=10^{-6}V$

Answer-(B)

The electric potential at a point

$P(x,y,z)$ is given by

$V = -x^2y -xz^3+4$ . The electric field

$\vec E$ at that point is

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$(2xy + z^3) \hat i + x^2 \hat j + 3xz^2 \hat k$
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$2xy \hat i + x^2 \hat j + 3xz^2 \hat k$
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$(2xy + z^3) \hat i + (x^2+y^2) \hat j + (3xy-y^2) \hat k$
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$2xy-z^3 \hat i + xy^2 \hat j + 3z^2x \hat k$

Explanation

$V=-x^2y-xz^3+4$

$\vec{E}=-\dfrac{\partial V}{\partial x}\hat{i}-\dfrac{\partial V}{\partial y}\hat {j}-\dfrac{\partial V}{\partial z}\hat{k}$

Now,

$\dfrac{\partial V}{\partial x}=-2xy-z^3$

$\dfrac{\partial V}{\partial y}=-x^2$

$\dfrac{\partial V}{\partial z}=-3xz^2$

Hence,

$\vec{E}=(2xy+z^3)\hat{i}+x^2\hat{j}+3xz^2\hat{k}$

Answer-(A)

The electric potential

$V$ at any point

$P(x,y,z)$ in space is given by

$V=4x^2\ V$ . The electric field at the point

$(1m,2m)$ is:

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$-8 \hat i$
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$8 \hat i$
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$-16 \hat i$
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$16 \hat i$

Explanation

$V=4x^2V$

$\vec{E}=-\dfrac{\partial V}{\partial x}\hat{i}-\dfrac{\partial V}{\partial y}\hat {j}-\dfrac{\partial V}{\partial z}\hat{k}$

Here,

$\dfrac{\partial V}{\partial x}=8x$

and,

$\dfrac{\partial V}{\partial y}=\dfrac{\partial V}{\partial z}=0$

Hence,

$\vec{E}=-8x\hat{i}$

Given point is

$(1,2)$ that is

$x=1,y=2$

Hence,

$\vec{E}=-8\hat{i}N/C$

Answer-(A)

The force experienced by a charged particle of charge

$q$ at

$(1,1,1)$ is

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$-5q \hat i$
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$5q \hat i$
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$5q (\hat i + \hat j)$
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$-5q (\hat i +\hat j)$

Explanation

$V=4x^2-3x$

$\vec{E}=-\dfrac{\partial V}{\partial x}\hat{i}-\dfrac{\partial V}{\partial y}\hat{j}-\dfrac{\partial V}{\partial z}\hat{k}$

$\dfrac{\partial V}{\partial x}=8x-3$

and,

$\dfrac{\partial V}{\partial y}=\dfrac{\partial V}{\partial z}=0$

Hence,

$\vec{E}=-(8x-3)\hat{i}$

At,

$(1,1,1):-$

$\vec{E}=-5\hat{i}N/C$

Hence, force

$\vec{F}=q\vec{E}$

$\implies \vec{F}=-5q\hat{i}N$

Answer-(A)

Electric potential due to a point charge

$q$ at a distance

$r$ from is given by

0%
$V = \dfrac{q}{4\pi\epsilon_0}$
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$V = \dfrac{q}{4\pi\epsilon_0r^2}$
0%
$V = \dfrac{q}{4\pi\epsilon_0r}$
0%
$V = \dfrac{q^2}{4\pi\epsilon_0r}$

Explanation

Electric potential at a point distant

$r$ from charge

$q$ is given by

$V=\dfrac{q}{4\pi\epsilon_o r}$

Answer-(C)

The value of

$^{(1,1,1)} _{(0,0,0)}\int E.dl$ is

0%
$-1V$
0%
$0V$
0%
$-2V$
0%
$1V$

Explanation

$V=4x^2-3x$

We know that

$\int Edl=-V$

Hence,

$\displaystyle\int_{(0,0,0)}^{(1,1,1)} Edl=-\left(V(1,1,1)-V(0,0,0)\right)$

$=-(1-0)$

$=-1V$

Answer-(A)

Find the total potential energy of the system.

0%
$5.48 \times 10^{-2}J$
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$5.48 \times 10^{2}J$
0%
$-5.48 \times 10^{2}J$
0%
$-3.6 \times 10^{-2}J$

Three charges

$Q, +q, -q$ are placed at the vertices of an isosceles right angled triangle as shown. The net electrostatic potential energy of the system is

$0$ if

$Q=$

0%
$-2q$
0%
$\dfrac{-q}{1+\sqrt2}$
0%
$2q$
0%
$\dfrac{-2q}{1+\sqrt2}$

Explanation

$V=V_{AB}+V_{AC}+V_{BC}$

$\implies V=\dfrac{1}{4\pi\epsilon_o}\left(\dfrac{q.q}{a}+\dfrac{qQ}{a}+\dfrac{qQ}{a\sqrt{2}}\right)=0$

$\implies \dfrac{q}{a}\left(q+Q+\dfrac{Q}{\sqrt{2}}\right)=0$

$\implies \left(\dfrac{\sqrt{2}q+\sqrt{2}Q+Q}{\sqrt{2}}\right)=0$

$\implies Q(1+\sqrt{2})=-\sqrt{2}q$

$\implies Q=\dfrac{-\sqrt{2}q}{1+\sqrt{2}}$

Answer-(D)

Charge

$5 \mu\ C$ ,

$-2 \mu\ C$ ,and

$-9 \mu\ C$ are placed at the comers A,B,C and D of a square ABCD of side

$2m$ . The electric potential at the center of the square is:

0%
$-27KV$
0%
$-27\sqrt { 2 }$
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$-90KV$
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$zero$

An infinite number of charges each equal to Q are placed along the x-axis at x=1,x=2,x=4,x=8,.......... and so on.Find the potential and electric filed at the point x=0 , due to this set of charges:

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$\dfrac{Q}{2 \pi \varepsilon_0},\dfrac{Q}{3 \pi \varepsilon_0}$
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$\dfrac{Q}{10 \pi \varepsilon_0},\dfrac{Q}{3 \pi \varepsilon_0}$
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$\dfrac{5Q}{ \pi \varepsilon_0},\dfrac{Q}{3 \pi \varepsilon_0}$
0%
$\dfrac{10Q}{2 \pi \varepsilon_0},\dfrac{Q}{3 \pi \varepsilon_0}$

Calculate potential energy of the system as shown in the figure

0%
$\dfrac{3kq^2}{d}$
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$\dfrac{\sqrt{3}kq^2}{d}$
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$\dfrac{2kq^2}{d}$
0%
$\dfrac{kq^2}{d}$

Find the speed of the proton at the final point.

0%
$2.7 \times 10^{-6}m/s$
0%
$3 \times 10^{8}m/s$
0%
$3 \times 10^{-8}m/s$
0%
$2.7 \times 10^{6}m/s$

Explanation

Mass of a proton is

$m_p = 1.67\times 10^{-27}\textrm{ kg}$

Charge of the proton is

$q = +1.602\times 10^{-19}\textrm{ C}$

Electric field is

$E = 8\times 10^4 \textrm{ N/C}$

Thus, the force experienced by proton is

$F = qE$

Acceleration is

$a = \displaystyle \frac{qE}{m_p}=\frac{1.602\times 10^{-19}\times 8\times 10^4}{1.67\times 10^{-27}} = 7.67\times 10^{12}\textrm{ m/s}^2$

Using the kinematics relation

$v^2-u^2 = 2a\Delta x$

where initial speed

$u=0$

$\implies v = \sqrt{2a\Delta x}$

Given : Displacement

$\Delta x = 0.5\textrm{ m}$

$\Rightarrow v = \sqrt{2\times 7.67\times 10^{12}\times 0.5} = 2.7\times 10^6\textrm{ m/s}$

Find the change in the potential energy of the proton in the displacement.

0%
$-6.4 \times 10^{-15}J$
0%
$6.4 \times 10^{-15}J$
0%
$-6.4 \times 10^{-19}J$
0%
$6.4 \times 10^{-19}J$

Explanation

Since the field is uniform, the potential difference will be given by the product of the electric field and the distance. so it will be

$8×0.5×10^4=4×10^4V$ Change in potential energy will be the work done by the proton.

That will be

$charge \times PD$

We know that one proton charge is

$1.6\times 10^{-19}C$

Hence the answer is

$-6.4 \times10^{-15}J$

The negative sign shows that the PE decreases as the proton does work.

Two charges

$q_1$ and

$q_2$ are placed

$30cm$ apart as shown. A third charge

$q_3$ is moved along the circle of radius

$40cm$ from

$C$ to

$D$ . The change in the potential energy of the system is

$\dfrac{q_3K}{4\pi\epsilon_0}$

Find K?

0%
$8q_2$
0%
$8q_1$
0%
$6q_2$
0%
$6q_1$

Explanation

Given two charges

$q_{1}$ and

$q_{2}$ are placed

$30\, cm$ apart as shown inn the figure. A third charge

$q_{3}$ moved from

$C$ to

$D$ .

Radius

$=40\, cm$

Given the change in potential energy of the system is

$\dfrac{q_{3}k}{4\pi \varepsilon _{0}}$ . We have to find the value of

$k$ .

We solve it as follows:

The change in potential energy of system is

$U_{D}-U_{C}$ as discussed under:

When charge

$q_{3}$ is at

$C$ , then its potential energy is

$U_{C}=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{q_{1}q_{3}}{0.4}+\dfrac{q_{2}q_{3}}{0.5} \right ]$

when charge

$q_{3}$ is at

$D$ , then

$U_{D}$ is

$U_{D}=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{q_{1}q_{3}}{0.4}+\dfrac{q_{2}q_{3}}{0.1} \right ]$

Hence, change in potential energy is

$\Delta U$

$\Delta U=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{q_{2}q_{3}}{0.1}-\dfrac{q_{2}q_{3}}{0.5} \right ]$ ...(i)

But given,

$\Delta U=\dfrac{q_{3}k}{4\pi \varepsilon _{0}}$ ...(ii)

On equating (i) and (ii), we get

$\dfrac{q_{3}k}{4\pi \varepsilon _{0}}=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{q_{2}q_{3}}{0.1}-\dfrac{q_{2}q_{3}}{0.5} \right ]$

We get,

$k=q_{2}(10-2)=8q_{2}$

Three charges

$Q,q,-q$ are placed at the vertices of an equilateral triangle. If the net electric potential energy of the system is

$0$ , then

$Q$ is

0%
$-q$
0%
$q$
0%
$0$
0%
None of the above

Explanation

$U=U_{AB}+U_{AC}+U_{BC}$

$\implies U=\dfrac{1}{4\pi\epsilon_o}\left(\dfrac{qQ}{l}+\dfrac{-qQ}{l}+\dfrac{-q.q}{l}\right)=0$

$\implies q(Q-Q-q)=0$

This gives both

$q=Q=0$

And, for any vlaue

$Q$ , potential energy of this system can't be

$0$ .

If three equal charges

$q$ are placed at the vertices of an equilateral triangle of side

$l$ , what is the net potential energy of the system?

0%
$\dfrac{3q^2}{4\pi\epsilon_0l}$
0%
$\dfrac{q^2}{2\pi\epsilon_0l}$
0%
$\dfrac{q^2}{4\pi\epsilon_0l}$
0%
$\dfrac{q^2}{\pi\epsilon_0l}$

Explanation

$U=\dfrac{q_1q_2}{4\pi\epsilon_o r}$

$U=U_{AB}+U_{BC}+U_{AC}$

$U=\dfrac{1}{4\pi\epsilon_o}\left(\dfrac{q^2}{l}+\dfrac{q^2}{l}+\dfrac{q^2}{l}\right)$

$\implies U=\dfrac{3q^2}{4\pi\epsilon_o l}$

Answer-(A)

An infinite nonconducting sheet has a surface charge density

$\sigma = 0.10\mu C/m^2$ on one side. How far apart are equipotential surfaces whose potentials differ by

$50 V$ ?

0%
$8.8mm$
0%
$8.8cm$
0%
$8.8m$
0%
$8.8\mu m$

Explanation

$E=\dfrac{\sigma}{2\epsilon_o}$

$V=El$

$\implies l=\dfrac{2V\epsilon_o}{\sigma}$

$\implies l=\dfrac{2\times 50\times 8.85\times 10^{-12}}{10^{-7}}$

$\implies l=8.85\times 10^{-3}m$

$\implies l=8.85mm$

Answer-(A)

The electric field in a region is given by :

$E=(4axy\sqrt z)\hat i + (2ax^2\sqrt z)\hat j + \dfrac{ax^2}{\sqrt z} \hat k$ , where

$a$ is a positive constant. The equation for an equipotential surface will be of the form

$k$ is a constant

0%
$z = \dfrac{k}{x^3y^2}$
0%
$z = \dfrac{k}{xy^2}$
0%
$z = \dfrac{k}{x^4y^2}$
0%
None of these

Two equal charges

$q$ are placed at a distance

$2a$ and a third charge

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

0%
$\dfrac{9q^2}{8\pi\epsilon_0a}$
0%
$\dfrac{q^2}{8\pi\epsilon_0a}$
0%
$\dfrac{-7q^2}{8\pi\epsilon_0a}$
0%
$\dfrac{6q^2}{8\pi\epsilon_0a}$

Explanation

Given two equal charges

$'q'$ are at a distance

$2a$ .

A third charge

$-2q$ is placed at the mid-point. We have to find the potential energy of the system.

Let

$O$ be the midpoint of the charges.

So, potential energy of the system is

$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{(-2q)(q)}{a}-\dfrac{(2q)(q)}{a}+\dfrac{(q)(q)}{2a} \right ]$

$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-2q^{2}}{a}-\dfrac{2q^{2}}{a}+\dfrac{q^{2}}{2a} \right ]$

$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-4q^{2}}{a}+\dfrac{q^{2}}{2a} \right ]$

$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-8q^{2}a+q^{2}a}{2a^{2}} \right ]$

$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-7q^{2}a}{2a^{2}} \right ]$

$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-7q^{2}}{2a} \right ]$

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

Two charges of equal magnitude '

$q$ ' are placed in air at a distance

$'2a'$ apart and third charge

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

$(\epsilon_0 =$ permittivity of free space) :

0%
$-\displaystyle\frac{q^2}{8\pi \epsilon_0 a}$
0%
$-\displaystyle\frac{3q^2}{8\pi \epsilon_0 a}$
0%
$-\displaystyle\frac{5q^2}{8\pi \epsilon_0 a}$
0%
$-\displaystyle\frac{7q^2}{8\pi \epsilon_0a}$

Explanation

Potential energy of the system

$U = \dfrac{-2q^2}{4\pi \epsilon_o a} + (\dfrac{-2q^2}{4\pi \epsilon_o a}) + \dfrac{q^2}{4\pi \epsilon_o (2a)}$

$\therefore$

$U = \dfrac{-2q^2}{4\pi \epsilon_o a} + (\dfrac{-2q^2}{4\pi \epsilon_o a}) + \dfrac{q^2}{4\pi \epsilon_o 2a}$

$U = \dfrac{q^2}{4\pi \epsilon_o a}\bigg[-2 -2 + \dfrac{1}{ 2}\bigg]$

$\implies$

$U = \dfrac{-7q^2}{8\pi \epsilon_o a}$

Capacity of a parallel plate capacitor is

$2\mu F$ . The two plates of the capacitor are given

$400\mu C$ and

$-200\mu C$ charges respectively. The potential difference between the plates is

0%
$100\ V$
0%
$200\ V$
0%
$300\ V$
0%
$150\ V$

Two parallel plate capacitors are connected in series. Each capacitor has a plate area A and a separation d between the plates. The dielectric constant of the medium between their plates are 2 and 4 . The separation between the plates of a single air capacitors of plate area A which effectively replaces the combination is:

0%
$\dfrac{2d}{3}$
0%
$\dfrac{3d}{2}$
0%
$\dfrac{3d}{4}$
0%
$\dfrac{8d}{5}$

A parallel plane capacitor

$C$ with plates of unit area and separation

$d$ is filled with a liquid of dielectric constant

$k=2$ . The level of liquid is

$\cfrac { d }{ 3 }$ initially. Suppose the liquid level decreases at a constant speed

$v$ , the time constant as a function of time

$t$ is:

0%
$\cfrac { 6\ { \varepsilon }_{ 0 }R }{ 5d+3vt }$
0%
$\cfrac { \left( 15d+9vt \right) { \varepsilon }_{ 0 }R }{ 2{ d }^{ 2 }-3dvt-9{ v }^{ 2 }{ t }^{ 2 } }$
0%
$\cfrac { 6\pi { \varepsilon }_{ 0 }R }{ 5d-3vt }$
0%
$\cfrac { \left( 15d-9vt \right) { \varepsilon }_{ 0 }R }{ 2{ d }^{ 2 }+3dvt-9{ v }^{ 2 }{ t }^{ 2 } }$

Explanation

At any time

$t$ , height of liquid will be

$d_1=d/3-vt$ and hence remaining height will be

$d_2=d-(\dfrac{d}{3}-vt)=\dfrac{2d}{3}+vt$

since capacitance

$C=\dfrac{\epsilon_0A}{d}$

$C_1=\dfrac{\epsilon_0}{2d/3+vt}$

$C_2=\dfrac{2\epsilon_0}{d/3-vt}$ , since

$A=1$ and

$k=1$

since both capacitor will act as connected in series ,

$\dfrac{1}{C}=\dfrac{1}{c_1}+\dfrac{1}{C_2}$

hence we get ,

$C=\dfrac{6\epsilon_0}{5d+3vt}$

and time constant is given by

$\tau=CR=\dfrac{6\epsilon_0R}{5d+3vt}$

Match the following two columns:

	Column I		Column II
A.	Electrical resistance	$1.$	$[ML^{3}T^{-3}A^{-2}]$
B.	Electrical potential	$2.$	$[ML^{2}T^{-3}A^{-2}]$
C.	Specific resistance	$3.$	$[ML^{2}T^{-3}A^{-1}]$
D.	Specific conductance	$4.$	None of these

0%
$A - 2, B - 3, C - 1, D - 4$
0%
$A - 2, B - 4, C - 3, D - 1$
0%
$A - 1, B - 2, C - 3, D - 3$
0%
$A - 1, B - 3, C - 2, D - 4$

Explanation

The dimensions of electrical resistance

$R = \dfrac {V}{I} = \dfrac {\left (\dfrac {W}{q}\right )}{I} = \dfrac {\dfrac {W}{It}}{(I)}$

$= \dfrac {W}{I^{2}t} = [ML^{2} T^{-2}T^{-1}A^{-2}]$

$= [ML^{2}T^{-3}A^{-2}]$
Then,

$(A) \rightarrow (2)$
The dimensions of electrical potential

$V = \dfrac {W}{q} = \dfrac {W}{It} = [ML^{2}T^{-2}A^{-1}T^{-1}]$

$= [ML^{2}T^{-3}A^{-1}]$
Then,

$(B)\rightarrow (3)$
The dimensions of specific resistance

$\rho = R \dfrac {l}{A} = [ML^{2}T^{-3}A^{-2}][L]$

$= [ML^{3}T^{-3}A^{-2}]$
Thus,

$(C) \rightarrow (2)$ and the dimensions of specific conductance

$\sigma = \dfrac {1}{\rho} = \dfrac {1}{[ML^{3}T^{-3}A^{-2}]}$

$= [M^{-1}L^{-3}T^{3}A^{2}]$

$=$ not give in column
Thus,

$(D) \rightarrow (4)$ .

A capacitor of capacitance

${ C }_{ 1 }=1\mu F$ can with stand maximum voltage

${ V }_{ 1 }=6kV$ (kilo-volt) and another capacitor of capacitance

${ C }_{ 2 }=3\mu F$ can withstand maximum voltage

${ V }_{ 2 }=4kV$ . When the two capacitors are connected in series, the combined system can withstand a maximum voltage of:

0%
$4kV$
0%
$6kV$
0%
$8kV$
0%
$10kV$

A capacitor of capacitance

$1\mu F$ withstands a maximum voltage of

$6\ kV$ , while another capacitor of capacitance

$2\mu F$ , the maximum voltage

$4\ kV$ . If they are connected in series, the combination can withstand a maximum of

0%
$6\ kV$
0%
$4\ kV$
0%
$10\ kV$
0%
$9\ kV$

The plates of a parallel plate capacitor are charged upto

$100 V$ .A

$2 mm$ thick insulator shunt is inserted between the plates . Then to maintain the same potential difference ,the distance between the same potential difference , the distance between the capacitor plates is increased by

$1.6 mm$ .The dielectric constant of the insulator is

0%
$5$
0%
$8$
0%
$4$
0%
$6$

Explanation

Here,

$V=100 V$ and

$t=2 mm$
As

$V=\dfrac{q}{c}$ and

$V$ and

$q$ remain unchanged ,
therefore C must be constant
ie,

$C_i=C_f$

$\dfrac{\varepsilon_0A}{d}=\dfrac{\varepsilon_0\,A}{d+1.6-t+\dfrac{t}{k}}$

$\therefore 1.6-t+\dfrac{t}{k}=0$
As

$t=2 mm$ ,therefore

$1.6-2+\dfrac{2}{k}=0$
or

$\dfrac{2}{k}=0.4$

$\therefore k=\dfrac{2}{0.4}=5$

A uniform electric field of magnitude

$290 V/m$ is directed in the positive

$x$ direction. A

$+13.0 \mu C$ charge moves from the origin to the point

$(x, y) = (20.0 cm, 50.0 cm).$

What is the change in the potential energy of the charge field system?

0%
$-754J$
0%
$-754mJ$
0%
$-754kJ$
0%
$-754\mu J$

Explanation

Given electric field,

$\overrightarrow{E}=290\, V/m$ directed along

$+x$ direction.

Charge

$=+13\mu C$ moves from origin to point

$(x,y)=(20\, cm , 50\, cm)$ .

We have to find the charge in potential energy of the charge field system.

We know, the change in potential energy

$=$ charge

$\times$ change in potential

i.e,

$\Delta U=q\Delta V$

But,

$\Delta V=-Ed$

$\Delta U=-qEd$

$=-13\times 10^{-6}\times 290\times 20\times 10^{-2}$

$=0.000754\, J$

$=-754\times 10^{-6}\, J$

Here

$d=20\, cm$ , since electric field is along positive

$x-$ axis.

A parallel plate capacitor of capacitance

$C$ is connected to a battery and is charge to a potential difference

$V$ . Another capacitor of capacitor

$2C$ is similarly charged to a potential difference

$2V$ . The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is:

0%
zero
0%
$\left( 3/2 \right) C{ V }^{ 2 }$
0%
$25/6C{ V }^{ 2 }$
0%
$\left( 9/2 \right) C{ V }^{ 2 }$

Explanation

Initial charges are given in this diagram now after joining, charges will flow due to potential difference.

Finally, the potential difference across the capacitors will be same as they are connected in parallel.

[ Potentntial difference across a capacitor

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

$\Rightarrow \dfrac{CV-q}{C}=\dfrac{-4CV+q}{2C}$

$\Rightarrow 2CV-2q=-4CV+q$

$\Rightarrow 6CV=3q$

$\Rightarrow q=2CV$

$\Rightarrow$ Total energy

$=\dfrac{\left(CV\right)^2}{2C}+\dfrac{\left(2CV\right)^2}{2C}$ [ Energy of a capactor

$=\dfrac{Q^2}{2C}$ ]

$=\dfrac{C^2V^2}{2C}+\dfrac{4C^2V^2}{2\left(2C\right)}$

$=\dfrac{CV^2}{2}=CV^2$

$=\dfrac{3CV^2}{2}$

$(B)$

Hence, the answer is

$\dfrac{3CV^2}{2}.$

The area of the plates of a parallel plate capacitor is

$A$ and the gap between them is

$d$ . The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance

$'y'$ from one plate as :

$K\ =\ \lambda \sec \left(\frac{\pi y}{2d}\right)$ , where

$\lambda$ is a dimensionless constant. The capacitance of this capacitor is

0%
$\large{\frac{\pi \epsilon_0 \lambda A}{2d}}$
0%
$\large{\frac{\pi \epsilon_0 \lambda A}{d}}$
0%
$\large{\frac{2\pi \epsilon_0 \lambda A}{d}}$
0%
None

Explanation

$E_y = \cfrac{\sigma}{ky\epsilon_0}$

$dV = \cfrac{\sigma}{k\epsilon_0} dy$

$dV = \cfrac{\sigma}{\lambda \epsilon_0} \cos {(\cfrac{\pi y}{2d})}dy$

Integrating to dy and other side;

$\int^{V}_{0} dV = \int^{d}_{0} \cfrac{\sigma}{\lambda \epsilon_0} \cos{(\cfrac{\pi y}{2d})} dy$

$V = \cfrac{\sigma}{\lambda \epsilon_0} . \cfrac{2d}{\pi}$

$C = \cfrac{Q}{\sigma}(\cfrac{\lambda \epsilon_0 \pi}{2d})$

$C = \cfrac{A \lambda \epsilon_0 \pi}{2d}$

In an

$LCR$ circuit as shown below both switches are open initially. Now switch

${S}_{1}$ is closed,

${S}_{2}$ kept open. (

$q$ is charge on the capacitor and

$\tau=RC$ is capacitive time constant). Which of the following statement is correct?

0%
At $t=\tau,q=CV/2$
0%
At $t=2\tau,q=CV(1-{e}^{-2})$
0%
At $t=\dfrac { \tau }{ 2 },q=CV(1-{e}^{-1})$
0%
Work done by the battery is half of the energy dissipated in the resistor.

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 13 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 13 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

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