MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 10

Two points charges

$4 \mu C$ and

$-2 \mu C$ are separated by a distance of 1 m in air. At what point in between the charges and on the line joining the charges, is the electric potential zero?

0%
In the middle of the two charges
0%
$1/3m$ from $4\mu C$
0%
$1/3m$ from $-2\mu C$
0%
Nowhere the potential is zero

Explanation

Let potential is zero at a distance

$x$ from

$4\mu c$

$\Rightarrow \dfrac{{k\left( 4 \right)}}{x} + \dfrac{{k\left( { - 2} \right)}}{{\left( {1 - x} \right)}} = 0$

$\Rightarrow \frac{4}{x} = \dfrac{2}{{\left( {1 - x} \right)}}$

$\Rightarrow 2 - 2x = x$

$\Rightarrow 2 = 3x$

$\Rightarrow x = \dfrac{2}{3}$

$\therefore$ from the distance is

$1 - \dfrac{2}{3} = \dfrac{1}{3}$

potential is zero at

$\dfrac{1}{3}m$ from

$- 2\mu c.$

Hence,

option

$(C)$ is correct answer.

1200950_1236059_ans_55bb658bc88a460294037b4d439dd60a.PNG

In a parallel plate capacitor, the region between the plates is filled by a delectric slab. The capacitor is connected to a cell and the slab is taken out.

0%
Some charge is drawn from the cell
0%
Some charge is returned to the cell
0%
The potential difference across the capacitor is reduced
0%
No work is done by an external agent in taking the slab out

Explanation

$\therefore$ Some charge is returned to the cell .

$\therefore$ Option

$B$ is correct answer.

1381153_1236957_ans_81cd73a335ac497db72effdd802b81e5.jpg

The distance between the plates of a parallel plate capacitor is

$d$ . A metal plate of thickness

$d/2$ is placed between the plates. The capacitance would be then be

0%
Unchanged
0%
Initial
0%
Zero
0%
Doubled

Explanation

$\begin{array}{l} Case\, I:\, \, \left( { Air\, \, between\, \, plates } \right) \\ Capaci\tan ce, \\ C=\dfrac { { { E_{ 0 } }A } }{ d } \\ Case\, \, II:\, \, \left( { Metal\, \, plate\, \, is\, \, inserted\, \, \, between\, \, the\, \, plates } \right) \\ capaci\tan e, \\ C'=\dfrac { { { E_{ 0 } }A } }{ { d-t+\dfrac { t }{ k } } } \\ Where\, \, \\ t=thickness\, \, of\, \, metal\, plate=\dfrac { d }{ 2 } \\ k=Dielectric\, \, cons\tan t\, \, of\, \, metal\, \, plate\, =\infty ] \\ C''=\dfrac { { { E_{ 0 } }A } }{ { d-t+\dfrac { { \dfrac { d }{ 2 } } }{ \infty } } } \\ =\dfrac { { { E_{ 0 } }A } }{ { \dfrac { d }{ 2 } } } \\ =\dfrac { { 2{ E_{ 0 } }A } }{ d } \\ =2C \\ Hence,\, it\, \, can\, \, be\, \, conculed\, \, that\, \, the\, \, new\, \, capaci\tan ce\, \, becomes\, \, \, double\, \, \, that\, \, of\, \, initial\, \, capaci\tan e. \end{array}$

A parallel plate capacitor consists of two circular plates each of radius 2 cm, separatrd by a distance of 0.1 mm If Voltage across the plates is at the rate of 5

$\times 10^{13}$ V/s, then the value of displacement current is

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

Explanation

$A=\pi 0^{2}$

$=4\pi \times 10^{-4}$

$I_{d}=\in _{0}A \left(\dfrac{dE}{dt}\right)$

$=8.85\times 10^{-12}\times 4\pi \times 10^{-4}\times 5\times 10^{13}$

$= 555.78\times 10^{-3}A$

Two identical particles of mass m carry a charge Q each . Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle a large distance with speed v. The closed distance of approach be

0%
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{mv}}$
0%
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{4{Q^2}}}{{m{v^2}}}$
0%
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{2{Q^2}}}{{m{v^2}}}$
0%
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{3{Q^2}}}{{m{v^2}}}$

Explanation

Due to repulsive force the other particle will start moving away. The velocity of the first particle will decrease while that of the other will increase. At the point of minimum distance between the two both the particles will be moving at same velocity. Let this velocity be

$u$

So using conservation of momentum we get

$mv=2mu$ or

$u=\dfrac{v}{2}$

The initial energy of the system is given as

$\dfrac{1}{2}mv^2$

And the energy at the minimum distance is given as

$\dfrac{1}{2}m\left(\dfrac{v}{2}\right)^2+\dfrac{1}{2}m\left(\dfrac{v}{2}\right)^2+\dfrac{1}{4\pi\epsilon_o}\dfrac{Q^2}{R}$

Equating the two energies we get

$\dfrac{1}{2}mv^2=\dfrac{1}{4}mv^2+\dfrac{1}{4\pi\epsilon_o}\dfrac{Q^2}{R}$

$\dfrac{1}{4}mv^2=\dfrac{1}{4\pi\epsilon_o}\dfrac{Q^2}{R}$

$R=\dfrac{1}{4\pi \epsilon_0}\dfrac{4Q^2}{mv^2}$

For high frequency a capacitor offer

0%
more reactance
0%
less reactance
0%
zero reactance
0%
inifinite reactance

Explanation

The relation between the reactance of capacitor and the frequency is given by

$X_C=\dfrac{1}{\omega C}$

$X_C=\dfrac{1}{2\pi f C}$

$X_c\propto \dfrac{1}{f}$

At high frequency, capacitor offer less reactance.

The correct option is B.

The plates of a parallel plate condenser are pulled apart with a velocity v. If at any instant their mutual distance of separation is x, the magnitude of the time of rate of change of capacity depends on x as follows :

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

Explanation

We know

$\displaystyle C=\frac{tA}{d}\Rightarrow \frac{dc}{dt}=tA\frac{d}{dx}(\frac{1}{x})$

$=\dfrac{-t_{0}A}{x^{2}}\left(\dfrac{dx}{dt}\right)$

Let

$d=x$

$= \dfrac{-t_{0}x}{x^{2}}\dfrac{dx}{dt}$

$\therefore \dfrac{dc}{dt}\propto \dfrac{1}{x^{2}}$

Equipotential surfaces are shown in fig, then the electric field strength will be

0%
100 Vm-1 along X-axis
0%
100 Vm-1 along Y-axis
0%
200 Vm-1 at an angle ${ 120 }^{ 0 }$ wirh X-axis
0%
50 Vm-1 at an angle ${ 180 }^{ 0 }$ wirh X-axis

Explanation

Using

$,$

$dV = - \overrightarrow E \cdot \overrightarrow {dr}$

$\Delta V = - E \cdot \Delta rcs\theta$

$E = \frac{{ - \Delta V}}{{\Delta r\cos \theta }}$

$E = \frac{{ - \left( { - 20 - 10} \right)}}{{10 \times {{10}^{ - 2}}\cos {{120}^0}}} = \frac{{ - 10}}{{10 \times {{10}^{ - 2}}\left( { - \sin {{30}^0}} \right)}} = \frac{{ - {{10}^2}}}{{ - 1/2}} = 200\,V/m$

direction of

$E$ be perpendicular to the equipotential surface

$i.e.,$ at

${120^0}$ with

$x-$ axis

$.$

Hence,

option

$(C)$ is correct answer.

The resultant capacitance between A and B in the following figure is equal to

0%
$1 \mu F$
0%
$3 \mu F$
0%
$2 \mu F$
0%
$1.5 \mu F$

P, Q and R are three points in a uniform electric field. The electric potential is

0%
minimum at R
0%
minimum at Q
0%
minimum at P
0%
Same at all three points

Explanation

We know that electric field is directed along decreasing potential.

$\therefore {V_Q} > {V_R} > {V_P}$

$\therefore$ electric potential is minimum at

$P$

hence, option

$C$ is correct.

A charged capacitor of capacitance C and having charge Q is to be connected with another uncharged capasitor of capasitance C' as shown till the steady state is reached , find the value of C' for heat liberated through the wires to be minimum.

0%
zero
0%
C
0%
C /2
0%
2C

Explanation

${u_i} = \frac{{{Q^2}}}{{2c}}$ ---------------

$(1)$

$c\left( {x - 0} \right) + \left( {x - 0} \right)\,c' = Q$

$\Rightarrow x\left( {c + c'} \right) = Q$

$\Rightarrow x = \frac{{Qc}}{{c + c'}}$

charge on

$c$

$=x c$

$=\frac{{Qc}}{{c + c'}}$

$\&$ charge on

$c' = \frac{{Qc'}}{{c + c'}}$

$\therefore {u_f} = \frac{{{Q^2}{c^2}}}{{{{\left( {c + c'} \right)}^2} \cdot 2c}} + \frac{{{Q^2}c{'^2}}}{{{{\left( {c + c'} \right)}^2} \cdot 2c'}}$

$= \frac{{{Q^2}}}{{2{{\left( {c + c'} \right)}^2}}}\left( {c + c'} \right) = \frac{{{Q^2}}}{{2\left( {c + c'} \right)}}$

$heat\, = {u_i} - {u_y} = \frac{{{Q^2}c'}}{{2\left( {c + c'} \right)}}$

$\therefore$ for heat to be min

$c' \to 0$

Hence,

option

$(A)$ is correct answer.

1215564_1263382_ans_2526bb7659094d9fb3a2e1d1a0e009b8.png

$75\%$ of the distance

$d$ between the parallel plates of a capacitor is filled with a meterial of dielectric constant

$K$ . Find the changed capacitance if original capacitance was

$C_{0}$ .....................

0%
$\left(\dfrac{3K}{K+3}\right) C_{0}$
0%
$\left(\dfrac{4K}{K+4}\right) C_{0}$
0%
$\left(\dfrac{4K}{K+3}\right) C_{0}$
0%
$\left(\dfrac{3K}{K+4}\right) C_{0}$

Maximum charge on capacitor after switch is closed is

0%
2 CE
0%
4 CE
0%
6 CE
0%
7 CE

Two capacitors of

$4\ \mu F$ and

$2\ \mu F$ are connected in series with the battery. If total potential difference across the two capacitors is

$200$ volts then the ratio of potential difference across one capacitor to another is

0%
$1:2$
0%
$2:1$
0%
$1:4$
0%
$4:1$

The capacitor is charged by closing the switch S. The witch is then opened and the capacitor is allowed to discharge. Take

${ R }_{ 1 }={ R }_{ 2 }={ R }_{ 3 }=R$ (Battery is ideal and connecting wire has negligible resistance). The fraction of the total heat generated, lost in

${ R }_{ 1 }$ during discharging is :

0%
$\dfrac { 1 }{ 6 }$
0%
$\dfrac { 1 }{ 3 }$
0%
$\dfrac { 1 }{ 2 }$
0%
$\dfrac { 2 }{ 3 }$

The ratio of charge densities on the surface of two conducting spheres is 3 :lithe radii of t: the spheres are 4 cm and 8 cm the ratio of the electric potential on the surfaces of the sphere 2 is

0%
3 : 4
0%
3 : 1
0%
1 : 3
0%
4 : 9

Explanation

Ratio of charge densities

$=3:2$ spheres are

$4cm$ and

$8cm$

spheres ratio

$=4:8=1:2$

${ E }_{ 1 }:{ E }_{ 2 }=?$

$E$

$\alpha$

$\dfrac { \lambda }{ r }$

$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } \times \dfrac { { r }_{ 2 } }{ { r }_{ 1 } }$

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

$=3:1$

${ E }_{ 2 }:{ E }_{ 1 }=1:3$

The energy per unit volume of a dielectric medium is proportional to square of

0%
relative permittivity
0%
charge
0%
energy
0%
electric intensity

Explanation

$E=\dfrac{V}{D}$ volt/meter

$=\dfrac{D}{k_q\varepsilon_o}$

$D=q/A$ ………

$(1)$

For energy

$dw=VCdv$

$w=C\displaystyle\int^{\forall}_oVdv=\dfrac{CV^2}{2}=\dfrac{1}{2}QV$ ……

$(2)$

from equation

$(1)$ &

$(2)$

$W=\dfrac{ED\cdot dA}{2}$

$\dfrac{ED(volume)}{2}$

$\dfrac{W}{V}\propto E$ .

1221070_1258664_ans_580369254b3243ed90ceb26f70acf43b.jpeg

The electric potential decreases uniformly from 120 V to 80 V as one moves on the X-axis from

$x = -1 cm$ to

$x = +1$ cm. The electric field at the origin.

0%
must be equal to $20 V/cm$
0%
must be equal to $2.0 V/cm$
0%
must be greater than $20 V/cm$
0%
must be less than $20 V/cm$

The distance between the plates of a parallel plate capacitor is 3 mm and the potential 1 difference applied is

$3\times10^5 V$ . If an electron travels from one plate to another, the change in its potential energy is

0%
$3\times10^5 eV$
0%
$900 eV$
0%
$10^8 eV$
0%
Negligible

Explanation

$v = 3 \times {10^5}v$

$d = 3 \times {10^{ - 3}}m$

$E = q\Delta v$

$= 1.6 \times {10^{ - 19}} \times 3 \times {10^5}$

$= 4.8 \times {10^{ - 14}}V$

$E = 3 \times {10^5}eV$

Hence,

option

$(A)$ is correct answer.

1216863_1270587_ans_3f04c1cab33f4306ab421d785c0f75f1.png

If electric intensity

$\overrightarrow { E }$ is along the X-axis, then the equipotential surfaces are parallel to

0%
XOY plane
0%
XOZ plane
0%
YOZ plane
0%
None of these

Explanation

Electric intensity

$\left( {\overrightarrow E } \right)$ is always perpendicular to equipotential surface

$.$ Since direction of electric intensity is along

$x-$ axis so equipotential plane will be

$YOZ$

HJence,

option

$(C)$ is correct answer.

Let

$V_0$ be the potential at the origin in an electric field

$\overset{\rightarrow}{E}=E_x\hat{i}+E_y\hat{j}$ . The potential at the point

$(x,y)$ is:

0%
$V_0-{_{x}{E}_x}-{_{y}{E}_y}$
0%
$V_0+{_{x}{E}_x}+{_{y}{E}_y}$
0%
${_{x}{E}_x}+{_{y}{E}_y}-V_0$
0%
$\left(\sqrt{x^2+y^2}\right)\sqrt{{E}_{x}^{2}+{E}_{y}^{2}}-V_0$

The electric potential at a distance of

$3 \,m$ on the axis of a short dipole of dipole moment

$4\times 10^{-12}$ coulomb-metre is

0%
$1.33 \times 10^{-3} V$
0%
$4 \,mV$
0%
$12 \,mV$
0%
$27 \,mV$

Explanation

$d=3m$

dipole moment

$=4\times { 10 }^{ -12 }cm$

$Q=\dfrac { P }{ d } =\dfrac { 4\times { 10 }^{ -12 }cm }{ 3m }$

$=1.33\times { 10 }^{ -12 }C$

The electric potential due to point charge at a point

0%
May be approximately zero
0%
May be positive
0%
May be negative
0%
Any of these

In order to increase the capacity of parallel plate condenser one should introduce between the plates, a sheet of

0%
mica
0%
tin
0%
copper
0%
stainless steel

Explanation

mica as it is having higher conductivity

$.$

$,$ increase the capacity of parallel plate condenser one should introduce between the plates

$,$ a sheet of

$mica.$

Hence,

option

$(A)$ is correct answer.

A parallel plate capacitor of area

$60 cm^2$ and separation 3 mm is charged initially to

$90 \mu C$ . If medium between the plates gets slightly conducting and the p!ate loses the charge initially at rate of

$2.5\times10^{-8} C/s$ , then what is the magnetic field between the plates?

0%
$2.5\times10^{-8} T$
0%
$2.0\times10^{-7} T$
0%
$1.63\times10^{-11} T$
0%
ZERO

If electric field in a region is zero, then electric potential in the region

0%
Must be zero
0%
Must not be zero
0%
May be zero
0%
None of these

Three capacitors 4, 6 and 12

$\mu F$ are connected in series to a 10 V source. The charge on the middle capacitor is

0%
10 $\mu C$
0%
20 $\mu C$
0%
60 $\mu C$
0%
5 $\mu C$

Explanation

$\begin{array}{l} \dfrac { 1 }{ { { C_{ eq } } } } =\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 6 } +\dfrac { 1 }{ { 12 } } \\ =\dfrac { { 3+2+1 } }{ { 12 } } \\ \Rightarrow { C_{ eq } }=2\, \mu F \\ Q=2\, \mu F\times 10 \\ =20\, \mu C. \end{array}$

Hence,

option

$(B)$ is correct answer.

In figure two points

$A$ and

$B$ are located in a region of electric field. The potential difference

$V_B-V_A$ is

0%
Positive
0%
Negative
0%
Zero
0%
None of these

Explanation

Electric field from

$A$ to

$B$

${ V }_{ B }-{ V }_{ A }=\left( -{ V }_{ e } \right)$ because, It goes to electric field always goes higher potential to lower potential. but,

${ V }_{ B }$ to

${ V }_{ A }$ is the opposite direction.

1241820_1293302_ans_76e3bcd50ce649818b3f590073377299.jpg

Two charges

$-5\mu C$ and

$+10\mu C$ are placed 20 cm apart. The net electric field at the mid-point between the two charge is

0%
$4.5\times { 10 }^{ 5 }N/C$ directed towards $+10\mu C$
0%
$13.5\times { 10 }^{ 5 }N/C$ directed towards $-5\mu C$
0%
$13.5\times { 10 }^{ 5 }N/C$ directed towards $+10\mu C$
0%
$4.5\times { 10 }^{ 5 }N/C$ directed towards $-5\mu C$

Explanation

R.E.F image

$\bar{E}net = \bar{E_{1}} + \bar{E_{2}}$

$= \frac{K(5 \times 10^{-6})}{10 \times 10^{-2}} + \frac{K(10 \times 10^{-6})}{10 \times 10^{-2}}$

$\Rightarrow 9\times 10^{9} [5 \times 10^{-5} + 10 \times 10^{-5}]$

$\Rightarrow 135 \times 10^{4}$ N\C

$\bar{E}net$

$13.5 \times$

$10^{5}$ N/C towards

$-5\mu c$

1239057_1301998_ans_1f25bf9339cb4b80a559f188ce788825.png

The change on capacitor

${ c }_{ 1 }$

0%
$0_{ \mu }c$
0%
$5_{ \mu }c$
0%
$10_{ \mu }c$
0%
none of these

In the following circuit the resultant capacitance between A and B is

$1\mu F$ . Find the value of C :

0%
$\dfrac { 23 }{ 32 } \mu F$
0%
$\dfrac { 32 }{ 23 } \mu F$
0%
$\dfrac { 13 }{ 23 } \mu F$
0%
$\dfrac { 23 }{ 13 }$

In the electric network shown, when no current flows through the

$4\, \Omega$ resistor in the arm

$EB$ , the potential difference between the points

$A$ and

$D$ will be :

0%
$3$ V
0%
$4$ V
0%
$5$ V
0%
$6$ V

Explanation

$-4+3+{ V }_{ ed }+0=0$

${ V }_{ ed }=1V$

${ V }_{ A }-{ V }_{ Q }=9-1=5V$

1380671_1302685_ans_7fca63eebc5c4c419f746324d15b2a76.jpg

The radius of the gold nucleus is

$6.6 \times {10^{ - 15}}m$ and the atomic number is

$79$ . The electric potential at the surface of the gold nucleus is :

0%
$1.7 \times {10^7}V$
0%
$7.1 \times {10^7}V$
0%
$1.7 \times {10^9}V$
0%
$7.1 \times {10^9}V$

Some charge Q is to be distributed on 3 concentric shell such that surface charge density on each shell is same as shown. Find the electric potential at r(<a) from point O.

0%
$\dfrac { Q(a+b-c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) }$
0%
$\dfrac { Q(a+b+c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) }$
0%
$\dfrac { Q(a+b+c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 }) }$
0%
$\dfrac { Q(a+b-c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }) }$

Explanation

$r<a$

from point

$0$ .

$\oint { \overline { E } } .d\overline { A } =\dfrac { { Q }_{ inside } }{ { \epsilon }_{ 0 } }$

${ \epsilon }_{ 0 }=8.854\times { 10 }^{ -12 }\dfrac { { e }^{ 2 } }{ { Nm }^{ 2 } }$

Now,

$E\times 4\pi { R }_{ 1 }^{ 2 }=\dfrac { -5\mu C }{ { \epsilon }_{ 0 } } =\dfrac { { Q }_{ inside } }{ { \epsilon }_{ 0 } }$

$E=\dfrac { { Q }_{ inside } }{ 4\pi { \epsilon }_{ 0 }{ R }_{ 1 }^{ 2 } }$

${ E }_{ 1 }=\dfrac { K{ Q }_{ incide } }{ { R }_{ 1 }^{ 2 } }$

${ E }_{ 2 }A=\dfrac { { Q }_{ inside } }{ { \epsilon }_{ 0 } }$

${ F }_{ y }4\pi { R }_{ z }^{ 2 }=\dfrac { Q }{ { \epsilon }_{ 0 } }$

Now,

$\dfrac { Q\left( a+b+c \right) }{ 4\pi { \epsilon }_{ 0 }\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }$

1244398_1347673_ans_8e468d1e6c594be8ad1e5df117c80592.jpg

The charges

$Q + q$ and

$+q$ are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, it the value of Q is:

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

Explanation

$U=K\left[\dfrac{q^2}{a}+\dfrac{Qq}{a}+\dfrac{Qq}{a\sqrt{2}}\right]=0$

$\Rightarrow q=-Q\left[1+\dfrac{1}{\sqrt{2}}\right]$

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

Two condenser of

$2 \mu F$ and

$4 \mu F$ are connected in series. The

$p.d$ of

$1200$ volt. The

$p.d$ across

$2 \mu F$ is -

0%
$400\ V$
0%
$600\ V$
0%
$800\ V$
0%
$900\ V$

Explanation

REF.Image

Q=CV

For a constant charge ,

$V \propto \frac{1}{C}$

$V_{A} \propto \frac{1}{C_{A}}$ &

$V_{B} \propto \frac{1}{C_{B}}$

$V_{A}=\frac{CB}{CA+CB}\times V$

$V_{A}=(\frac{4}{2+4})\times 1200$

$= \frac{4}{6}\times 1200$

= 800 V

1166244_1359331_ans_134f50da30e246c89176351549dbddb0.jpg

Potential difference between two points is equal to

0%
electric charge /time
0%
work done/time
0%
work done/charge
0%
work done $\times$ charge

Explanation

Potential difference between two points is equal to work done / charge.

$V=\dfrac { W }{ q }$

or,

$W=V.q$ (this is the working formula).

Two capacitors 'A' of

$3 \mu F$ and 'B' of

$2\mu F$ are connected in series. 'A' can withstand a potential difference of 3 KV while 'B' can withstand 5 KV. What maximum potential difference can be applied across the combination?

0%
$8 KV$
0%
$6 KV$
0%
$7.5 KV$
0%
$8.3 KV$

The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure:
What is the value of current at

$t = 4s$ ?

0%
$3\ \mu A$
0%
$2\ \mu A$
0%
$Zero$
0%
$1.5\ \mu A$

Explanation

Since

$\dfrac{dq}{dt}|_{t = 4s} = 0$

$\therefore current = 0$

The energy stored in a parallel plate capacitor can be treated as the energy filled. The energy per unit volume, due to the electric field is :

0%
$E^2$
0%
$\dfrac{1}{2}{\epsilon_0}{E^2}$
0%
$\dfrac{1}{{2{\epsilon_0}}}{E^2}$
0%
$\frac{1}{2}{E^2}$

The potential energy of a 1

$\mathrm { kg }$ particle free to move alongthe

$x$ -axis is given by

$V ( x ) = \left( \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right) J$

The total mechanical energy of the particle is

$21 .$ Then, the maximum speed (in ms

$^ { 1 }$ ) is

0%
$\dfrac { 3 } { \sqrt { 2 } }$
0%
$\sqrt { 2 }$
0%
$\dfrac { 1 } { \sqrt { 2 } }$
0%
2

Explanation

$\begin{array}{l} TE=PE+KE \\ for\, \, KE\, \, \to mass \\ PE=0 \\ \Rightarrow KE=2J \\ \Rightarrow \dfrac { 1 }{ 2 } m{ v^{ 2 } }=2 \\ \therefore v=2\, m/s \end{array}$

Option

$D$ is correct.

An infinite number of charges 'q' each are placed along the x - axis at x =1, x=4 ,x=8 and so on. If the distance are in meters calculate the electric potential at x=0

0%
$\frac{{3q}}{{8\pi {E_0}}}$
0%
$\frac { q } { 2 \pi \epsilon _ { 0 } }$
0%
$\frac { 2 q } { \pi \epsilon _ { 0 } }$
0%
$\frac { 4 q } { \pi \epsilon _ { 0 } }$

Explanation

$\begin{array}{l} { V_{ 0 } }=\frac { { kq } }{ 1 } +\frac { { kq } }{ 4 } +\frac { { kq } }{ 8 } +....................... \\ =kq\left( { 1+\frac { 1 }{ { { 2^{ 2 } } } } +\frac { 1 }{ { { 2^{ 3 } } } } +.......... } \right) \\ =kq\left( { 1+\frac { { \frac { 1 }{ 4 } } }{ { 1-\frac { 1 }{ 2 } } } } \right) \\ =kq\left( { 1+\frac { 2 }{ 4 } } \right) \\ =\frac { { 3kq } }{ 2 } =\frac { { 3q } }{ { 8\pi { E_{ 0 } } } } \end{array}$

Hence, Option

$A$ is correct.

In the situation shown in figure, what should be the relation between Q and q so that electric potential at centre of the square is zero:

0%
$Q=q$
0%
$Q=3q$
0%
$Q=2q$
0%
$Q=-3q$

Explanation

$Q=$ Sum motion of three side total charges

$=q+q+q$

$=3q$

1380708_1330482_ans_979cb73ef7bd420bb1663c917af965e9.jpg

The figure shows an experiment plot for discharging of a capacitor in an

$R-C$ circuit. The time constant

$t$ of this circuit lies between

0%
$0\ and\ 50\ sec$
0%
$50\ and\ 100\ sec$
0%
$100\ and\ 150\ sec$
0%
$150\ and\ 200\ sec$

Explanation

During discharging of a capacitor

$V = {V}_{0}{e}^{\frac{–t}{\tau}}$ where

$t$ is the time constant of RC circuit.

$t =\tau$ ,

$V =\dfrac{{V}_{0}}{e}=0.37{V}_{0}$

From the graph,

$t = 0$ ,

${V}_{0}= 25$ V

$\therefore V = 0.37 \times 25$ V

$= 9.25$ V

This voltage occurs at time lies between

$100$ sec and

$500$ sec.

Hence, time constant

$\tau$ of this circuit lies between

$100$ sec and

$500$ sec.

Two spheres of capacitance

$3\mu F$ and

$5\ mu F$ are charged to

$300\ V$ and

$500\ V$ respectibely and are connected together. The common potential in steady state will be :

0%
$400\ V$
0%
$425\ V$
0%
$350\ V$
0%
$375\ V$

In the capacitor shown in the circuit is charged to 5V and left in the circuit,in 12s the charge on the capacitor will become.

0%
$\frac {10}{e} C$
0%
$\frac {e}{10} C$
0%
$\frac {10}{e^2} C$
0%
$\frac {e^2}{10} C$

The electric potential at a point

$(x,0,0)$ is given by

$V=\left[\dfrac{1000}{\chi }+\dfrac{1500}{\chi }+\dfrac{500}{\chi }^3\right]$
then the electric field at

$x =1\, m$ is (in volt/m)

0%
$-5500 \, \hat{i}$
0%
$5500 \, \hat{i}$
0%
$\sqrt{5500 \, \hat{i}}$
0%
Zero

Potential difference between the points B and C of the circuit is

0%
$\dfrac { ( C_2-C_1 ) }{V}$
0%
$\dfrac { ( C_4-C_3 ) }{V}$
0%
$\dfrac { ( C_2C_3 - C_1C_4 ) }{ (C_1+C_2+C_3+C_4) } V$
0%
$\dfrac {C_1C_4 - C_2C_3 }{ ( C_1+C_2) \times ( C_3+C_4) } V$

An arc of radius r carries change. The linear density of charge is

$\lambda$ and the arc subtends a angle

$\frac{\pi }{3}$ at the center. What is electric potential at the center

0%
$\dfrac{\lambda }{4\varepsilon _{0}}$
0%
$\dfrac{\lambda }{8\varepsilon _{0}}$
0%
$\dfrac{\lambda }{12\varepsilon _{0}}$
0%
$\dfrac{\lambda }{16\varepsilon _{0}}$

Explanation

An arc of radius

$=r$

linear density of charge

$=\lambda$

subtends a angle

$=\dfrac { \pi }{ 3 }$ the centre.

$V=\dfrac { 1 }{ 4 } \dfrac { \lambda }{ 12{ \epsilon }_{ 0 } }$

$V=\dfrac { 1 }{ 4 } \times \dfrac { \lambda }{ 3{ \epsilon }_{ 0 } } =\dfrac { \lambda }{ 12{ \epsilon }_{ 0 } }$

The potential at a point, due to a positive charge of

$10\mu { C }$ at a distance of

$9 m$ is

0%
$10^{ { 5 } }V$
0%
$10^{ { 3 } }V$
0%
$10^{ { 6 } }V$
0%
$10^{ { 4 } }V$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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