MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 9

Two capacitors of

$1\mu F$ and

$2\mu F$ are connected in series and this combination is changed upto a potential difference of

$120$ volt. What will be the potential difference across

$1 \mu F$ capacitor:

0%
$40 volt$
0%
$60 volt$
0%
$80 volt$
0%
$120 volt$

Explanation

Given,

$c_1=1\mu f,c_2=2\mu f,PD=120v$

$C_{eq}=\dfrac{c_1c_2}{c_1+c_2}=\dfrac{2\times1}{2+1}=\dfrac{2}{3}\mu f$

We know,

$Q=cv$ Where Q is the charge, C is the capacitance of the capacitor and v is the potential difference.

Now,

$Q_{net}$ in circuit is equivalent capacitance of capacitors attached in the circuits is multiplied by PD

$Q_{net}=12\times\dfrac{2}{3}=80$

$Q=cv=1\mu fv=80\Rightarrow v=80v$

An AC source producing emf

$\epsilon = \epsilon_0[ cos(100 \pi S^{-1})t + cos (500 \pi s^{-1})t]$ is connected in series with a capacitor and a resister. The steady-state current in the circuit is found to be:

$i = i_1 [cos(100 \pi s^{-1})t + \phi_1] + i_1 cos[(500 \pi s^{-1})t + \phi_1]$

0%
$i_1 > i_2$
0%
$i_1 = i_2$
0%
$i_1 < i_2$
0%
the information is insufficient to find the relation between $i_1$ and $i_2$ .

Two small spheres have mass

${m}_{1}$ and

${m}_{2}$ and hanging from massless insulating threads of lengths

${l}_{1}$ and

${l}_{2}$ . Two spheres carry charges

${q}_{1}$ and

${q}_{2}$ respectively. The spheres hang such that they are on the same horizontal level and the threads are inclined to the vertical at angle

${\theta}_{1}$ and

${\theta}_{2}$ respectively. If

$F_1 = F_2$ , then:

0%
${ \theta }_{ 1 }={ \theta }_{ 2 }$
0%
${ M }_{ 1 }={ M }_{ 2 }$
0%
$\cfrac { l_{ 1 } }{ \tan { { \theta }_{ 1 } } } =\cfrac { l_{ 2 } }{ \tan { { \theta }_{ 2 } } } \quad$
0%
$\cfrac { q_{ 1 } }{ \tan { { \theta }_{ 1 } } } =\cfrac { q_{ 2 } }{ \tan { { \theta }_{ 2 } } }$

Explanation

For sphere

$1$

In equilibrium, from figure.

$\begin{array}{l} { T_{ 1 } }\cos { \theta _{ 1 } } ={ M_{ 1 } }g; \\ { T_{ 1 } }\sin { \theta _{ 1 } } ={ F_{ 1 } } \\ \therefore \tan { \theta _{ 1 } } =\dfrac { { { F_{ 1 } } } }{ { { M_{ 1 } }g } } . \end{array}$

For sphere

$2$

In equilibrium, from figure.

$\begin{array}{l} { T_{ 2 } }\cos { \theta _{ 2 } } ={ M_{ 2 } }g; \\ { T_{ 2 } }\sin { \theta _{ 2 } } ={ F_{ 2 } } \\ \therefore \tan { \theta _{ 2 } } =\dfrac { { { F_{ 2 } } } }{ { { M_{ 2 } }g } } . \end{array}$

Force of repulsion between two charges are same

$\therefore F_1=F_2$

$\theta_1=\theta_2$ only if

$\dfrac{{{F_1}}}{{{M_1}g}} = \dfrac{{{F_2}}}{{{M_2}g}}.$

But

$F_1=F_2$ , then

$M_1=M_2$ .

A capacitor of capacitance

$C_1$ charged at a certain potential v. It is connected with another uncharged capacitor

$C_2$ . What is the final p.d. of this new system.

0%
$\dfrac{C_2 V}{C_1 C_2}$
0%
$\dfrac{C_1 V}{C_1 C_2}$
0%
$(1 + \dfrac{C_2}{C_1})$
0%
$D(1 - \dfrac{C_2}{C_1}) V$

Explanation

Charge on first capacitors,

${q_1} = {c_1}v$

Charge on second capacitor,

${q_2} = c$

when they are connected in parallel the total charge,

$q = {q_1} + {q_2}$

$\therefore {q_1} = {c_1}v$

and capacitor,

$c = {c_1} + {c_2}$

Let

$V$ be the common potential difference across each capacitors, then

$q=cv$

$\therefore v' =\large \frac{q}{c} = \frac{{{c_1}v}}{{{c_1} + {c_2}}}$

On changing a capacitor with charge Q stored energy is W. If charge is doubled then stored energy will be:-

0%
$2W$
0%
$4W$
0%
$8W$
0%
$\dfrac{1}{2}W$

Explanation

$\begin{array}{l} w=\frac { Q }{ { 2C } } \\ { w^{ ' } }=\frac { { 4Q } }{ { 2C } } \\ =4w \\ Hence, \\ option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$

The capacitors A and B are connected in series with a battery as shown in the figure. When the switch S is closed and the two capacitors get charged fully, then-

0%
The potential difference across the plates of A is $4V$ and across the plates of B is $6V$
0%
The potential difference across the plates of A is $6V$ and across the plates of B is $4V$
0%
The ratio of electric energies stored in A and B is 2:3
0%
The ratio of charges on A and B is 3:2

Explanation

Given,

Both capacitors are in series

Capacitance of A & B, ${{C}_{A}}=2\mu F\,\,\And \,\,{{C}_{B}}=3\mu F$

Total Potential is ${{V}_{T}}=({{V}_{A}}+{{V}_{B}})=10V$

Let,

Potential difference on capacitor A & B, ${{V}_{A}},\,{{V}_{B}}$

In series, Capacitor have equal charge

${{Q}_{A}}={{Q}_{B}}$

${{C}_{A}}{{V}_{A}}={{C}_{B}}{{V}_{B}}$

$\dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}}{{{C}_{A}}}$

$\dfrac{{{V}_{A}}+{{V}_{B}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}+{{C}_{A}}}{{{C}_{A}}}$

${{V}_{B}}=\dfrac{{{C}_{A}}({{V}_{A}}+{{V}_{B}})}{{{C}_{B}}+{{C}_{A}}}=\dfrac{2\times 10}{5}=4V$

${{V}_{B}}=4V$

Similarly, $\dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}}{{{C}_{A}}}\Rightarrow \,\,\,{{V}_{A}}=\dfrac{{{V}_{B}}\times {{C}_{B}}}{{{C}_{A}}}\,\,$

$\Rightarrow \,{{V}_{A}}=\dfrac{4\times 3}{2}=6V$

Potential difference on capacitor A & B, ${{V}_{A}}=6V,\,\,\,\,{{V}_{B}}=4V$

An uncharged metal object

$M$ is insultated from its surroundings.A positively charges metal sphere

$S$ is then brought near to

$M$ . Which diagram illustrate the resultant distribution of charge on

$S$ and

$M$

Explanation

Since all negative charges attracted towards

$S$ so the move towards left and leaving positive charges on right.
1174540_1068892_ans_c7fb05572a104cdab28fe9da73f8279d.png

The electrostatics potential inside a charged spherical is given by

$\phi=ar^2+b$ where

$r$ is the difference between centre

$a,b$ are constants. Then the charge density of the ball is

0%
$-6a\varepsilon_or$
0%
$-24\pi a\varepsilon_o$
0%
$-6a\varepsilon_o$
0%
$-24\pi a \varepsilon_or$

Explanation

$E=-\cfrac{d\phi}{dr}$ ;

$E=-2ar$

and

$E=\cfrac{1}{4\pi\epsilon_o}\cfrac{q}{r^2}=-2ar$

$q=-4\pi\epsilon_o\cdot 2ar^3$

Charge density,

$\rho=\cfrac{q}{Volume}=\cfrac{-4\pi\epsilon_o2ar^3}{\cfrac{4}{3}\pi r^3}$

$\rho=-6a\epsilon_o$

The capacity of a parallel plate condenser is

$5 \mu F$ .When a glass plate is placed between the plates of the condenser,its potential difference reduces to 1/8 of original value.The magnitude of relative dielectric constant of glass is

0%
2
0%
6
0%
7
0%
8

Explanation

Potential of a capacitor without dielectric,

$V=\dfrac{Q}{C}=\dfrac{Qd}{A \varepsilon_0}$ where

$Q$ is the charge on the capacitor

$d$ is the plate distance

$A$ is the area of plates

$\varepsilon_0$ is permeability of free space

After insertion of dielectric,potential becomes:

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

$\dfrac{Qd}{kA\varepsilon_0}$ where k is dielectric constant.

since,

$V'=\dfrac{V}{8}$

$\dfrac{Qd}{kA\varepsilon_0}=\dfrac{Qd}{8A \varepsilon_0}$

$k=8$

In the given figure, the capacitance of each capacitor is C, then the effective capacitance between A and B will be

0%
$\frac3{C}{2}$
0%
C
0%
12C
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none of these

A parallel plate capacitor of capacitance

$6\mu F$ in air and

$60\mu F$ . When dielectric is introduced. What is the dielectric constant of the medium.?

0%
$0.4$
0%
$0.1$
0%
$0.77$
0%
$0.25$

Explanation

Capacitance in air,

$C=6\mu F$

Capacitance in Dielectric,

$c'=60\mu F$

Capacitance of parallel plate capacitor is increased by insertion of a dielectric medium by '

$k$ ' times.

where

$k=$ Dielectric constant

$C = \dfrac {E_{0}}{D}$ ,

$C' = k \dfrac{E_{0}}{D}$

$\therefore c'=kC$

$k=\dfrac{C}{c'}$

$k=\dfrac{6}{60}=0.1$ .

In the circuit below, if a dielectric is inserted into

$C_{2}$ then the charge on

$C_{1}$ will

0%
Increase
0%
Decrease
0%
Remain same
0%
Be halved

Explanation

Initially wire between both capacitors is neutral, as the source is connected. Positive charge moves toward B and negative charge move toward A.

when the dielectric is placed between the capacitor plate charge increase, due to which wire ends B having more positive charge flow toward it and more positive charge flow toward B.

Hence, charge on capacitor ${{C}_{1}}$ also increase

A series combination of two capacitances of value

$0.1\ mu F$ and

$1\mu F$ is connected with a source of voltage

$500\ volts$ . The potential difference in volts across the capacitor of value

$0.1\ muF$ will be :

0%
$50$
0%
$500$
0%
$45.5$
0%
$454.5$

Explanation

Given,

Capacitance, ${{C}_{1}}=0.1\,\mu F\,\,and\,\,{{C}_{2}}=1\,\mu F$

In series charge is equal

$Q={{C}_{1}}{{V}_{1}}={{C}_{2}}{{V}_{2}}$

${{V}_{2}}=\dfrac{{{C}_{1}}{{V}_{1}}}{{{C}_{2}}}$

In series total potential difference is sum of all paternal difference

$V={{V}_{1}}+{{V}_{2}}$

$V={{V}_{1}}+\dfrac{{{C}_{1}}{{V}_{1}}}{{{C}_{2}}}={{V}_{1}}\left( \dfrac{{{C}_{2}}+{{C}_{1}}}{{{C}_{2}}} \right)$

${{V}_{1}}=\dfrac{{{C}_{2}}V}{{{C}_{2}}+{{C}_{1}}}=\dfrac{1\times 500}{1+0.1}=454.54\,V$

Hence, Potential difference across $0.1\,\mu F\,\,\,is\,\,\,454.5\,V$

The equivalent capacitance between the point s

$A$ and

$B$ in the given diagram is

0%
$8\ \mu F$
0%
$6\ \mu F$
0%
$\dfrac {8}{3}\ \mu F$
0%
$\dfrac {3}{8}\ \mu F$

An Uncharged capacitor of capacitance

$C$ is connected to a battery of emf

$\varepsilon$ at

$t=0$ through a resistance

$R$ , then
(i) the maximum rate at which energy is stored in the capacitor is:

0%
$\dfrac {\varepsilon^2 }{ 4R }$
0%
$\dfrac {\varepsilon^2 }{2R}$
0%
$\dfrac {\varepsilon^2} {R}$
0%
$\dfrac {2 \varepsilon^2}{R}$

Explanation

The rate of growth of charge for the capacitor

$q=\varepsilon C\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)$

Let E be the energy stored inside the capacitor.

Then,

$E=\dfrac{{{q}^{2}}}{2C}$

$E=\dfrac{{{\varepsilon }^{2}}{{C}^{2}}}{2C}{{\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)}^{2}}$

$E=\dfrac{{{\varepsilon }^{2}}C}{2}\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)$

Let r be the rate of energy stored inside the capacitor.

Then,

$r=\dfrac{dE}{dt}$

$r=\dfrac{2{{\varepsilon }^{2}}C}{2}\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)\left( -{{e}^{\dfrac{-t}{RC}}} \right)\left( \frac{-1}{RC} \right)$

$r=\dfrac{{{\varepsilon }^{2}}}{R}\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)\left( {{e}^{\dfrac{-t}{RC}}} \right)$

Now, at $t = 0$

The rate of energy stored is

$r=\dfrac{{{\varepsilon }^{2}}}{R}$

Hence, the rate of energy stored inside the capacitor is $\dfrac{{{\varepsilon }^{2}}}{R}$

Choose the incorrect relation among the following:

0%
Ampere = $\frac{coulomb}{second}$
0%
Volt = $\frac{coulomb}{joule}$
0%
ohm = $\frac{Volt}{Ampere}$
0%
$1 coulomb = 6.25 \times 10^{18} electron charge$

Explanation

Ampere =

$SI$ unit of current

$=\dfrac{Charge (Coulomb)}{Time (Second)}$

Volt =

$SI$ unit of Potential

$=\dfrac{Work (Joule)}{Charge (Coulomb)}$

Ohm =

$SI$ unit of Resistance

$=\dfrac{Potential (Volt)}{Current (Ampere)}$

$1\ Coulomb = (1.6\times 10^{-19})^{-1}$ electron charge

$=6.25\times 10^{18}$ electron charge

Option

$B$ is incorrect.

For spherical symmetrical charge distribution, variation of electric potential with distance from centre is given in diagram. Given that:-

$V = \dfrac{q}{4 \pi \varepsilon_0 R_0}$ for

$r \ R_0$ and

$V = \dfrac{q}{4 \pi \varepsilon_0 r}$ for

$r \geq R_0$ .
Then which option is incorrect:

0%
Total charge within $2 R_0$ is $q$ .
0%
Total electrostatic energy for $r \leq R_0$ is zero
0%
At $r = R_0$ electric field is discontinuous.
0%
There will be no charge anywhere except at $r = R_0$

Explanation

$\begin{array}{l} V=\dfrac { q }{ { 4\pi { \varepsilon _{ o } }{ R_{ o } } } } \\ \therefore che\, \, is\, \, q \\ So\, \, option\, \, A\, \, is\, \, correct\, \\ E=0\, \, \, inside\, \, r={ R_{ o } } \\ =\frac { { Kq } }{ { { R^{ 2 } } } } \, \, \, \, \, outside\, \, r\ge { R_{ o } } \\ \therefore E\, \, is\, \, discontinuous\, \, of\, \, r={ R_{ o } } \\ So\, \, option\, \, C\, \, is\, \, correct \\ also\, \, there\, \, will\, \, be\, \, no\, \, che\, \, except\, \, at\, \, r={ R_{ o } } \\ So\, \, option\, \, D\, \, is\, correct. \\ electrostatics\, \, energy\, \, for\, \, r<{ R_{ o } } \\ is\, \, zero\, \, but\, \, if\, \, is\, \, not\, \, zero\, \, for\, \, r={ R_{ o } } \\ So\, \, option\, \, B\, \, is\, \, wrong \\ Hence, \\ option\, \, A,C\, \, and\, D\, \, is\, \, correct\, \, answer. \end{array}$

in a certain region of free space a non-uniform electric field which depends on x-coordinates is given by

$\vec E = E_0 x \hat i$ . What is the total amount of electric potential energy contained within the cube of sides of length L as shown below?

0%
$\dfrac{1}{5} \varepsilon_0 E_0^2 L^5$
0%
$\dfrac{1}{5} \varepsilon_0 E_0^2 L^6$
0%
$\dfrac{1}{6} \varepsilon_0 E_0^2 L^6$
0%
$\dfrac{1}{2} \varepsilon_0 E_0^2 L^5$

Explanation

$\begin{array}{l} from\, the\, given\, figure \\ du=\frac { 1 }{ 2 } { \varepsilon _{ 0 } }\left( { { t^{ 2 } } } \right) .\left( { { 1^{ 2 } }dx } \right) \\ =\frac { 1 }{ 2 } { \varepsilon _{ o } }{ L^{ 2 } }E_{ 0 }^{ 2 }\int _{ 0 }^{ L }{ { x^{ 2 } }dx } \\ =\frac { 1 }{ 2 } { \varepsilon _{ 0 } }{ L^{ 2 } }E_{ 0 }^{ 2 }\left( { \frac { { { L^{ 3 } } } }{ 3 } } \right) \\ =\frac { 1 }{ 2 } { \varepsilon _{ 0 } }{ L^{ 5 } }E_{ 0 }^{ 2 }. \\ Hence,\, the\, option\, D\, is\, the\, \, correct\, answer. \end{array}$

In the circuit shown in figure

$C=6\mu F$ . The charge stored in the capacitor of capacity

$C$ is(

$Voltage =10V$ ):

0%
Zero
0%
$90\mu C$
0%
$40\mu C$
0%
$60\mu C$

In the given circuit the potential at point

$E$ is:

0%
Zero
0%
$-8V$
0%
$-\dfrac{4}{3}V$
0%
$\dfrac{4}{3}V$

Four charges

$1$ mc,

$2$ mc,

$3$ mc, ...

$6$ mc are placed on at corner of a square of side

$1$ m. The square lies in XY plane with its centre at origin

0%
The electric potential at origin at origin
0%
The electric potential is zero every where along X axis
0%
The electric potential is not zero along Z- axis
0%
The electric potential is zero along Z- axis for any orientation of square in XY plane.

Explanation

Electric field at origin= sum of electric field due to individual charges

Then,

option

$C$ is correct because electric field is not zero along z-axis as a every point the potential equals sum of potential due to individual charges which can never be zero.

1228596_1125647_ans_b9b0cb13576b47689e2cbb974ee65257.PNG

The equivalent capacitance of the combination shown in figure is:

0%
$C$
0%
$2C$
0%
$C/2$
0%
none of these

A charge of 10 e.s.u is placed at the distance of 2 cm from a charge of 40 e.s.u and 4 cm from another charge of 20 e.s.u.The potential energy of charged 10 e.s.u is?

0%
87.5
0%
112.5
0%
150
0%
250

Explanation

Potential energy of system of Two charge

$U=k\dfrac{Q_{1}Q_{2}}{r}(s.2)$

$=\dfrac{k \ Q_{1}Q_{2}}{e} (C.G.S) \quad k=\dfrac{1}{4\pi \epsilon_{0} }$

Potential energy at

$10 \ esu$ charge in

$U= \dfrac{10\times 40}{2}+\dfrac{10\times 20}{4}= 250 \ erg$

1382649_1131313_ans_87d78080150a4a23ba8643648a0d54da.png

A parallel plate capacitor consists of two circular disc separated by a small distance it is capacitance of see the radius of discus half and separation between disc is doubled a medium of dielectric constant K is introduced between the desk the new capacitance is found to be same as that of the original capacitance C the dielectric constant of medium is

0%
$2$
0%
$3$
0%
$8$
0%
$6$

Find the potential difference across point a and b in figure.

0%
$(\dfrac{\varepsilon _1 + \varepsilon _2}{c_1 + C_2})C_1$
0%
$(\dfrac{\varepsilon _1 + \varepsilon _2}{2c_1 + C_2})C_1$
0%
$(\dfrac{\varepsilon _1 + \varepsilon _2}{3c_1 + C_2})C_1$
0%
$(\dfrac{\varepsilon _1 + \varepsilon _2}{4c_1 + C_2})C_1$

Explanation

${V_1} = {Q \over {{C_1}}},{V_2} = {Q \over {{C_2}}}$

$- {\varepsilon _1} + {V_1} - {\varepsilon _2} + {V_2} = 0$

$\left( {{\varepsilon _1} + {\varepsilon _2}} \right) = {V_1} + {V_2}$

${\varepsilon _1} + {\varepsilon _2} = {Q \over {{C_1}}} + {Q \over {{C_2}}}$

$Q = \left( {{\varepsilon _1} + {\varepsilon _2}} \right)\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$

${V_2} = {Q \over {{C_2}}}$

${V_2} = \left( {{\varepsilon _1} + {\varepsilon _2}} \right)\left( {{{{C_1}} \over {{C_1} + {C_2}}}} \right)$

1095423_1115917_ans_a08590299ee74224874c30c23ab25446.png

The charge after flow through circuit is after switch is closed

0%
ZERO
0%
10C
0%
5C
0%
30C

Negative pole of capacitor is situated at point

$x = 0$ and positive plate is placed at

$x = 3d$ . A dielectric slab of thickness

$d$ is introduced in capacitor. The dielectric slab is at equal distance from these plates. When we move from point

$x = 0$ to

$x = 3d$ , then:-

0%
intensity of electric field remains same
0%
initially electric potential increases, then decreases and finally increases
0%
electric potential increases continuously
0%
direction of electric field is not changed.

Two plates of equal area A , one a square of side

$2L \times 2L$ and other a rectangle os sides

$4L \times L$ are used to construct a parallel plate capacitor with plates separation . d . if C is the capacitance , then C=

0%
$\frac { \varepsilon _{ 0 }A }{ 2d }$
0%
$\frac { \varepsilon _{ 0 }A }{ d }$
0%
$\frac {2 \varepsilon _{ 0 }A }{ d }$
0%
none of these

The capacitance of a parallel plate capacitor is

$2.5\mu F$ . When it is half-filled with a dielectric shown in the figure Its capacitance becomes

$5\mu F$ , the dielectric constant of the dielectric is :

0%
$7.5$
0%
$3.0$
0%
$0.33$
0%
$4.0$

Explanation

Let area of each plate be A and spacing between then be d.

Then capacitance of parallel plate capacitor,

$C = \dfrac{\epsilon_0A}{d} = 2.5\mu F$

When it is half filled with dielectric, capacitance of capacitor filled with air,

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

capacitance of capacitor filled with medium of dielectric constant

$k$ is

$C_2 = \dfrac{k\epsilon_0 (A/2)}{d}$

Since

$C_1$ and

$C_2$ are in parallel, equivalent capacitance will be,

$C' = C_1+C_2 = 5\mu F$

$\dfrac{\epsilon_0A}{2d} +\dfrac{k\epsilon_0}{2d} = 5$

$\dfrac{2.5}{2} +\dfrac{k\times2.5}{2} = 5$

$1+k = 4$

$k =3$

A parallel plate capacitor with air between the plates has a capacitance of 2pF. The capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 is

0%
12pF
0%
24pF
0%
6pF
0%
18pF

Explanation

We know,

$C=\dfrac{Ak\varepsilon_0}{d}$

Given,

For

$k=1,$

$C=\dfrac{A\varepsilon_0}{d}=2$

For k=6, and

$d\rightarrow \dfrac{d}{2}$

$C=\dfrac{A\times 6\varepsilon_0}{\dfrac{d}{2}}$

$C=\dfrac{12A\varepsilon_0}{d}=12C=24pF$

Option

$\textbf B$ is the correct answer

A dielectric slab is inserted between the plates of a capacitor is

$Q$ and the magnitude of the induced charges on each surface of the dielectric is

$Q$ ,

0%
$Q^{'}$ may be larger then $Q$
0%
$Q^{'}$ may be smaller then $Q$
0%
$Q^{'}$ may be equal then $Q$
0%
None of the above

As shown the figure an insulating material of dielectric constant K is inserted into of the space between the plates.If initial capacitance of the capacitor was C, then its new capacitance will be

0%
$\dfrac{1}{2}C(K +1)$
0%
$\dfrac{1}{2}$ $\dfrac{C}{1 + K}$
0%
$\dfrac{1}{2}$ $\dfrac{1 + K}{C}$
0%
$C(1 + k)$

Six identical square metallic plates are arranged as shown in figure length of each plate is l the capacitance of this arrangement should be

0%
$3\epsilon_0l^2/d$
0%
$4\epsilon_0l^2/d$
0%
$3\epsilon_0l^2/2d$
0%
$2\epsilon_0l^2/d$

An infinite ladder of capacitors, each

$1\mu F$ , is made as shown in figure. The capacitance between A and B (in

$\mu F$ ) is

0%
$1$
0%
$1.6$
0%
$3.4$
0%
$0$

Explanation

From the figure,

$\dfrac{X}{X+1}+1=X$

$X+X+1=X(X+1)$

$2X+1=X^2+X$

$X+1=X^2$

$X^2-X-1=0$

$X=\dfrac{1+\sqrt{5}}{2}$

$X=\dfrac{1+2.236}{2}=1.6\mu F$

1474705_1134356_ans_cd6cdecc45cd443cb3c6bc66ac53d401.jpeg

A capacitor of capacitance 1

$\mu F$ withstands the maximum voltage 6 kV while a capacitor of 2

$\mu F$ withstands the maximum voltage 4 kV. What maximum voltage will the system of these two capacitor withstands if they are connected in series?

0%
10 kV
0%
12 kV
0%
8 kV
0%
9 kV

An electron in a picture tube of TV set is accelerated from rest through a potential difference of

$5\times 10^3V$ .Then the speed of electron as a result of acceleration is going to be

0%
$1.2 \times 10^7m/s$
0%
$2.2\times 10^7m/s$
0%
$3.2 \times 10^7m/s$
0%
$4.2\times 10^7m/s$ .

Explanation

By energy conservation

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

$\dfrac{1}{2}mv^2=1.6\times 10^{-19}\times 5\times 10^3$

$v^2=\dfrac{2\times 1.6\times 10^{19}\times 5\times 10^3}{9.1\times 10^{-31}}$

$v^2=0.175\times 10^{16}$

$v=0.419\times 10^8$

$=4.19\times 10^{7}$

$=4.2\times 10^7m/s$ .

The potential difference between points

$A$ and

$B$ in the circuit shown in figure, will be:-

0%
$1 V$
0%
$2 V$
0%
$-3 V$
0%
None of these

The radius of a nucleus of an atom

$(Z=50)$ is

$9\times 10^{-10}m$ then potential on its surface, will be:

0%
$80V$
0%
$8kV$
0%
$9V$
0%
$9kV$

Explanation

$R=9\times 10^{-10}m$

$q=charge=ze$

$q=50\times 1.6\times 10^{-19}C$

$V=potential =\dfrac{kq}{R}$

$V=\dfrac{9\times 10^9\times 50\times 1.6\times 10^{-19}}{9\times 10^{-10}}$

$V=80$ Volt

A circuit
has a section AB as shown in figure with E= ${\text{10}}\;\;{\text{V,}}\;{{\text{C}}_{{\text{1}}\;}}{\text{ = }}$ 1.0 $\mu F$ ${{\text{C}}_{\text{2}}}\;{\text{ = }}\;{\text{2}}{\text{.0}}$ $\mu F$ and the potential difference ${{\text{V}}_{\text{A}}}\;{\text{ - }}{{\text{V}}_{\text{B}}}{\text{ = 5}}\;{\text{V}}$ The
voltage across ${{\text{C}}_{\text{1}}}$ is :

0%
Zero
0%
5 V
0%
10 V
0%
15 V

A parallel plate capacitor C is charged by connecting it to a battery using a switch S as shown in the figure, Now S is opened and the plate separation is then increased. As a result:

0%
the charged stored in C increase
0%
the energy stored decreases
0%
the P.D. between the plates increases
0%
the intensity of the electric field between the increase

During charging and discharging of a capacitor:-

0%
Current flows in the circuit, which is constant during charging or discharging duration
0%
No current flows in the circuit
0%
Current flows in the circuit and is varying with time
0%
During chargung current is constant but while discharging current is variable

Explanation

During charging of capacitor, the charge on capacitor increases and hence voltage across capacitor increases as,

$Q = CV$ . Due to increase in voltage, the flow of current increases.

Similarly, during dicharging of capacitor, the charge on capacitor decreases and hence voltage across capacitor decreases. Due to decrease in voltage, the flow of current also decreases.

So current flows in circuits and is varying with time.

Two identical capacitors are connected in series with a source of potential V. If Q is the charge on one of the capacitors, the capacitance of each capacitor is:

0%
Q/2V
0%
2Q/V
0%
Q/V
0%
None of these

Explanation

In series connection charge on each capacitor would be constant also equivalent capacitance in series

$c'=\dfrac{C}{2}$ [ following

$\dfrac{1}{c'}=\dfrac{1}{c_1}+\dfrac{1}{c_2}$ ] and voltage

$V$ is applied across it so,from capacitive law,

$Q=c'v \Rightarrow Q=\dfrac{CV}{2} \Rightarrow C=\dfrac{2Q}{V}$

$2 \mu F$ capacitance has P.D across its two terminals of 200 V. It is disconnected from the battery and when another uncharged capacitance is connected in parallel to it, P.D becomes 20 V. The capacity of another capacitance will be:

0%
$2\mu F$
0%
$4 \mu F$
0%
$10\ \mu F$
0%
$16 \mu F$

Explanation

$Q = 2 \times 200 = 400 \,mu \,C$

$V_f = 20 \,V$

$q = CV_f = 20 \,C$

$Q - q = 2V_f = 40 \,\mu C$

$20 \,C + 40 = Q = 4\omega \,mu C$

$C = 10 \,\mu F$
1055722_1159939_ans_021dfbaefbdc4643a918066c60d518e4.1159939-sol

What is the electric potential at a distance

$'x'$ from the centre, inside a conducting sphere having a charge

$Q$ and radius

$R$ ?

0%
$\dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q}{R}$
0%
$\dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q}{x}$
0%
$\dfrac {1}{4\pi \epsilon_{0}} \dfrac {QX}{R^{2}}$
0%
Zero

At any point on the perpendicular bisector of the line joining two equal and opposite charges.

0%
The electric field is zero
0%
The electric potential is zero
0%
The electric potential decreases with increasing distance from their mid point
0%
The electric field is perpendicular to the line joining the charges

An electric circuit requires a total capacitance of

$2 \mu F$ across potential of

$1000 V$ . Large number

$1 \mu F$ capacitances are available each of which would breakdown if the potential is more than

$350 V$ How many capacitances are required to make the circuit.

0%
$24$
0%
$20$
0%
$18$
0%
$12$

Explanation

$\begin{array}{l} { C_{ total } }=2\mu F \\ v=1000v \\ Number\, \, of\, \, capacitor\, \, in\, series \\ =\dfrac { { 1000 } }{ { 30 } } \\ \simeq 3\left( { 2.8 } \right) \\ { C_{ series } }=\dfrac { 1 }{ 3 } \\ Now, \\ 2=P\times \dfrac { 1 }{ 3 } \\ P=6\left( { in\, \, parallel } \right) \\ total\, \, number\, \, of\, capaci\tan ce\, \\ =3\times 6 \\ =18 \\ \therefore \, option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$
1215219_1208122_ans_f9ac1b78c7834a7fa1554e5c89691b95.JPG

1215219_1208122_ans_f9ac1b78c7834a7fa1554e5c89691b95.JPG

The capacitance of a parallel plate condenser does not depend upon

0%
the distance between the plates
0%
area of the plates
0%
medium between the plates
0%
metal of the plates

Two capacitors of caacity

${ C }_{ 1 }$ and

${ C }_{ 2}$ are connected in series and potential difference V is applied across it. Then the potential difference across

${ C }_{ 1 }$ will be

0%
$V\frac { { C }_{ 2 } }{ { C }_{ 1 } }$
0%
$V\frac { { C }_{ 1 }+{ C }_{ 2 } }{ { C }_{ 1 } }$
0%
$V\frac { { C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } }$
0%
$V\frac { { C }_{ 1 } }{ { C }_{ 1 }+{ C }_{ 2 } }$

Explanation

$Given:$ two capacitors of capacity

$C_1$ and

$C_2$ are connected in series and potential difference V is applied across it.

$Solution:$ We know that,

$Q=C_{\text {net }} V$

$Q=\left(\dfrac{C_{1} C_{2}}{c_{1}+C_{2}}\right) V$

$\therefore$

$,$ Voltage across

$C_{1}=\dfrac{Q}{C_{1}}$

$=\dfrac{\dfrac{\left(c_{1} c_{2}\right) v}{(\left. c_{1}+c_{2}\right)}}{c_{1}}$

$=\dfrac{c_{2}}{c_{1}+c_{2}} v$

$So,the$

$correct$

$option:C$

2010499_1190358_ans_7f87f38973a84200a2a9048476236d13.jpeg

Two identical parallel plate capacitors of same dimensions are connected to a D.C. source in series When one of the plates of one capacitor is brought closer to the other plate

0%
the voltage on the capacitor whose plates came closer is greater than the voltage on the capacitor whose plates are not moved.
0%
the voltage on the capacitor whose plates came closer is smaller than the voltage on the capacitor whose plates are not moved.
0%
the voltages on the two capacitors remain equal.
0%
the applied voltage is divided equally between the two capacitors

Explanation

Capacitance is given by,

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

where

$A =$ Area of plates,

$d =$ distance between the plates

If one plate is brought closed, i.e,

$d$ decreases, capacitance of that capacitor will increases.

Since,

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

If capacitance of capacitor is increased, Voltage of that capacitor will decreases.

So, Voltage on the capacitor whose plates came closer is smaller than the voltage on the capacitor that is not moved.

The capacitor is charged to equilibrium. A dielectric of constant

$K$ is slowly inserted in the capacitor. Find work done by the external agent in inserting the dielectric.

0%
$\dfrac{1}{2} c\varepsilon^2 (k - 1)$
0%
$\dfrac{1}{2} c\varepsilon^2 (1 - k)$
0%
$\dfrac{1}{2} kc\varepsilon^2$
0%
$-\dfrac{1}{2}kc\varepsilon^2$

Explanation

$\dfrac{1}{2}c\varepsilon^2; \dfrac{1}{2} kc\varepsilon^2$

$\Delta U = \dfrac{1}{2} c\varepsilon^2 (k - 1)$

$= -\Delta U$

$= \dfrac{1}{2} c\varepsilon^2 (1 - k)$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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