MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 2

Two charges

$+q$ and

$-q$ are kept apart. Then at any point on the right bisector of line joining the two charges

0%
The electric field strength is zero
0%
The electric potential is zero
0%
Both electric potential and electric field strength are zero
0%
Both electric potential and electric field strength are non-zero

Explanation

At equatorial point

${ E }_{ e }=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { p }{ { r }^{ 3 } }$
(directed from

$+q$ to

$-q$ ) and

${ V }_{ e }=0$

The intensity of electric field E due to charge Q at distance r.

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$E\propto r$
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$E\propto \displaystyle\frac{1}{r^2}$
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$E\propto \displaystyle\frac{1}{r}$
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$E\propto \displaystyle\frac{1}{r^3}$

Explanation

According to Coulomb's Law:

$E = k \frac{Q}{r^2}$
where E is electric field, q is charge at a distance r and k is a constant. Therefore,

$E \propto \frac{1}{r^2}$

Identify the dimension of electric potential,

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$L^{2}MT^{-3}I^{-1}$
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$L^{1}MT^{-2}I^{-1}$
0%
$L^{-2}MT^{-2}I^{-1}$
0%
none of the above

Explanation

Work done

$=$ charge

$\times$ potential difference

or Force

$\times$ distance

$=$ (current\times time)

$\times$ potential difference

$[MLT^{-2}]\times [L]=([IT])[V]$

$[V]=[ML^2T^{-3}I^{-1}]$

Two charges are separated by a distance d. If the distance between them is doubled, how does the electric potential between them change?

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It is doubled
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It is halved
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It is quartered
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It is quadrupled
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It is unchanged

Explanation

Initial electric potential between the charges

$V= \dfrac{kq}{d}$

New distance between the charges

$d' = 2d$

Thus new electric potential

$V' = \dfrac{kq}{2d} =\dfrac{V}{2}$

Capacitance is best described as the:

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Rate of flow of charge
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Energy per unit charge
0%
Change in electric potential per unit length
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Charge per unit of electric potential difference
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Electric potential difference per unit of electric current

Explanation

Capacitance is defined as the charge stored in the capacitor per unit electric potential difference applied across it.

$\therefore$ Capacitance

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

The diagram shows equipotential lines from an unknown charge configuration. Determine where in the diagram, the electric field is the strongest :

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A
0%
B
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C
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D

Explanation

The electric field is given by

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

Hence electric field is greatest at place with highest potential gradient, that is, where the equipotential surfaces are most dense, which is at point C.

When moving electron comes closer to other stationary electron, then its kinetic energy and potential energy respectively _____ and _____.

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Increases, increases
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Increases, decreases
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Decreases, increases
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Decreases, decreases

The dielectric constant of air is.

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$8.85\times 10^{-12}C^2N^{-1}M^{-2}$
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$1$
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Infinite
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Zero

Explanation

Dielectric constant of a medium

$K = \dfrac{\epsilon}{\epsilon_o}$

where

$\epsilon$ is the permittivity of the medium.

For air,

$\epsilon = \epsilon_o$

$\implies$

$K = 1$

A method for charging a conductor without bringing a charged body in contact with it is called:

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Magnetization
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Electrification
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Electrostatic induction
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Electromagnetic induction

The retarding potential is generally greater in magnitude than the acceleration potential.

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True
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False

What is not true of equipotential surface?

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The PD between any two points on the surface is zero
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The electric field is always perpendicular to the surface
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Equipotential surfaces are always spherical
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No work is done in moving a charge along the surface

Explanation

Equipotential surface

$\rightarrow$ P.D difference between two points on the surface is zero always since potential is same everywhere in equipotential surface.

$\rightarrow$ The EF is always perpendicular to the surface because there is no potential gradient along any direction parallel to the surface P so no EF parallel to the surface

$\rightarrow$ Equipotential surface can have any shape not just sphere.

$\rightarrow$ No work is done in moving a charge along the surface, because potential difference is zero.

Hence option (C) is correct

A parallel plate capacitor is charged. If the plates are pulled apart

0%
The capacitance increases
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The potential difference increases
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The total charge increases
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The charge and potential difference remain the same

What is the unit of electric potential difference?

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Volt
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Coulamb
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Joul
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Watt

The equivalent capacitance between the points A and B in the given diagram is:

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

Explanation

${C_S}$ Capacitance in series

$\begin{array}{l} \dfrac { 1 }{ { { C_{ S } } } } =\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \\ =\dfrac { 3 }{ 2 } \\ { C_{ S } }=\dfrac { 2 }{ 3 } \end{array}$

The equivalent capacitance between A & B is

$\begin{array}{l} { C_{ P } }=\dfrac { 2 }{ 3 } +2 \\ { C_{ P } }=\dfrac { 8 }{ 3 } \mu F \end{array}$

1103641_1023766_ans_1d5714d7ffb6408099834fd2bcf4458c.png

If the susceptibility of dia, para and ferro magnetic materials are

$Xd. Xp. Xf$ respectively, then

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$Xd < Xp < Xf$
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$Xf < Xp< Xd$
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$Xf < Xd < Xp$
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$Xd < Xf < Xp$

The potential of a sphere of radius

$2\ cm$ when a charge of

$2\ coulomb$ is given to it, will be

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$9\times 10^{3}V$
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$9\times 10^{11}V$
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$9\times 10^{6}V$
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$9\times 10^{16}V$

Calculate the electrostatic potential energy of two electrons separated bt 3

$\overset { \circ }{ A }$ in vacuum

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$4.81eV$
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$5.81eV$
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$6.71eV$
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$7.82eV$

Out of the given statements which of the following are true :

A) Work done in moving a charge on equipotential surface is zero.

B) Electric lines of force are always normal to equipotential surface.

C) When two like charges are brought closer, the electrostatic potential energy of the system is decreased.

D) Electric lines of force converge at positive charge and diverge at negative charge.

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A, B, C, D are true
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A, B, C are true
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A, B are true
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A only true

Explanation

A) Work done by electric field = q

$\times$ Potential difference
For an equipotential surface, change in potential is zero. Thus, work done is zero.
B) Electric lines of force are always normal to an equipotential surface because if the field has any tangential component then it would cause motion of the charges present on the surface contradicting its equipotential property.
C) Bringing two like charges of the system increases the potential energy of the system (because like charges always repel).
D) Electric lines of force start at positive charges and end at negative charges.

When a dielectric material is introduced between the plates of a charged condenser, after disconnecting the battery, the electric field between the plates:

0%
decreases
0%
increases
0%
does not change
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may increase or decrease

Explanation

We know electric field between capacitor plates:

$E= \dfrac{\sigma}{\epsilon _0}$

$= \dfrac{V}{d}$ , where

$d$ is the distance between the plates.

When battery is disconnected, the change present on capacitor is fixed, i.e.,

$V$ is constant.

Now, when the dielectric is introduced, the electric field becomes:

$E' = \dfrac{V}{Kd}$ , where K is dielectric constant which is always > 1

$\therefore$ electric field decreases.

A condenser is charged and then battery is removed. A dielectric plate is put between the plates of condenser, then correct statement is

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Q is constant, V and U decrease
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Q is constant, V increases U decreases
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Q increases, V decreases U increases
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Q, V and U increase

Explanation

$\textbf{Step 1 - Given data Refer Figure}$

let Cross-section of plate

$= A$

distance between plate

$= d$

surface charge density

$= \sigma$

Capacitance of capacitor

$= c$

$\textbf{Step 2 - Explanation of condenser without slab}$

Since, we know that - when battery is removed from a charge capacitor

then charge is constant.

So, electric field inside capacitor

$= E = \dfrac {\sigma}{\varepsilon_{0}}$

Potential between plate

$= V = E \cdot d = \dfrac {\sigma d}{\varepsilon_{0}}$

$Capacitance = c = \dfrac {Q}{v} = \dfrac {\sigma A}{v} = \dfrac {\sigma A}{\left (\dfrac {\sigma d}{\varepsilon_{0}}\right )}$

$c = \dfrac {A\varepsilon_{0}}{d}$

energy store in capacitance

$= v = \dfrac {1}{2} \dfrac {\theta^{2}}{c}$

$\textbf{Step 3 - Explanation of condenser with dielectric slab}$

Now, when we interduce a di-electric in capacitor then -

electric field between plate

$= E' = \dfrac {E}{k} = \dfrac {\sigma}{k\varepsilon_{0}}$

Potential between plate

$= v' = E'\cdot d = \dfrac {\sigma d}{k\varepsilon_{0}} = \dfrac {v}{k}$

Capacitance of capacitor

$= c' = \dfrac {\theta}{v'} = \dfrac {\sigma A}{\left (\dfrac {\sigma d}{k\varepsilon_{0}}\right )}$

$e' = \dfrac {k\varepsilon_{0}A}{d} = ck$

So, store energy -

$v' = \dfrac {1}{2} \dfrac {\theta^{2}}{c'} = \dfrac {1}{2} \dfrac {\theta^{2}}{ck} = \dfrac {v}{k}$

$\textbf{Step 4 - Observation -}$

$\bullet$ Since, we can see that - after inducing the slab in capacitor is -

$Capacitance = c' = ck \rightarrow increase$

$Potential = v' = \dfrac {v}{k} \rightarrow decrease$

$energy = u' = \dfrac {u}{k} \rightarrow decrease$

$\rightarrow$ Hence option (A) is correct.

Two charges

$q$ and

$-q$ are kept apart. Then at any point on the perpendicular bisector of line joining the two charges:

0%
the electric field strength is zero
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the electric potential is zero
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both electric potential and electric field strength are zero
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both electric potential and electric field strength are non-zero

Explanation

We know voltage is given by

$\dfrac{Kq}{r}$
All the points on perpendicular bisectore are at equidistant from the two points A and B.

$\therefore Potential = \dfrac{Kq}{r}+\dfrac{K(-q)}{r}$

$= 0$

Statement(A): Negative charges always move from a higher potential to lower potential point

Statement (B): Electric potential is vector.

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A is true but B is false
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B is true but A is false
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Both A and B false
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Both A and R are true

A parallel plate condenser is charged by connecting it to a battery. Without disconnecting the battery, the space between the plates is completely filled with a medium of dielectric constant k. Then:

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potential becomes 1/k times
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charge becomes k times
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energy becomes 1/k times
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electric intensity becomes k times

A parallel plate capacitor, filled with a material of dielectric constant

$K$ , is charged to a certain voltage and is isolated. The dielectric material is removed. Then:
(a) capacitance decreases by a factor

$K$
(b) electric field reduces by a factor

$K$
(c) voltage across the capacitor increases by a factor

$K$
(d) charge stored in the capacitor increases by a factor

$K$

0%
(a) and (b) are true
0%
(a) and (c) are true
0%
(b) and (c) are true
0%
(b) and (d) are true

Explanation

Initial capacitance

$c= \dfrac {k\varepsilon_0 A}{d}$
When the system is isolated, charge (

$Q$ ) present on it is constant.
Now, when dielectric is removed

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

$\Rightarrow$

$C$ decreases by a factor of

$K$ .

We know that

$Q=C \times V$

$C$ decreases by factor of

$K$ and

$Q$ remains constant,

$V$ increases by a factor of

$K$ .

Twenty seven identical mercury drops each charged to $10V$ , are allowed to form a big drop. The potential of the big drop is

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$90 V$
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$9 V$
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$900 V$
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$270 V$

Explanation

Voltage of each mercury drop

$= 10 V$

$\Rightarrow \dfrac {kq}{r}= 10 V$

$\Rightarrow q= \dfrac {10r}{k}$

$total \ charge =\dfrac {27\times 10r}{k}$
Equating volumes to find radius of bigger charge,

$\dfrac {4}{3}\pi (R^3)=27\cdot \dfrac {4}{3}\pi r^3$

$R=3r$
final voltage

$= \dfrac {kq}{R}=\dfrac {k \times 27\times 10r}{k\times 3r}$

$=90 V$

(1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery.

(2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.

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Both 1 and 2 are true, 2 is not correct explanation of 1
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Both 1 and 2 are true, 2 is correct explanation of 1.
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1 is false, 2 is true
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1 is true, 2 is false

Explanation

Consider a capacitor with charge density

$\sigma$ .
The potential between its two plates is given by

$\sigma d/ \epsilon_{0}$
When a dielectric is inserted, the electric field inside the capacitor decrease decreasing the potential between the two plates of capacitor.
However, this is nothing to the dielectric strength of the dielectric.

One plate of parallel plate capacitor is smaller than the other. The charge on the smaller plate:

0%
will be less than other
0%
will be more than other
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will be equal to other
0%
will depend upon the medium between the plates

Explanation

Hint: apply gauss law for the two plates (conductors) where E=0.

Step 1: imagine a Gaussian surface

Imagine a Gaussian surface surrounding the two plates as shown in figure.

Step 2: Apply gauss law

Gauss law: electric flux through any closed surface is directly

proportional to the charge enclosed in it.

$\oint{E.dA}$ = $\dfrac{q}{{{\varepsilon _o}}}$

As both plates are conductor hence E=0 for the Gaussian surface

$\dfrac{q}{{{\varepsilon _o}}}$ = 0

${\varepsilon _o} \ne 0$

So q=0

Step 3: find charge on each plate

As, total q=0

So charge on each plate will be equal and opposite

Final step: The charge on the smaller plate will be equal to other plate to make total charge zero on both the plates irrespective of shape and size.

A dielectric of thickness $5$ cm and a dielectric constant $10$ is introduced between the plates of a parallel plate capacitor having plate area $500 sq.$ cm and separation between the plates $10cm$ . The capacitance of the capacitor with the dielectric slab is $\varepsilon _{0}=8.8\times 10^{-12}C^{2}/N-m^{2}$

0%
$4.4 pF$
0%
$6.2 pF$
0%
$8 pF$
0%
$10 pF$

Explanation

We know

$C=\dfrac{kl_{0}A}{d}$
effective capacitors is series combination of

$\dfrac{kl_{0}A}{5\ cm}\ and\ \dfrac{l_{0}A}{5\ cm}$

$\therefore \dfrac{1}{C_{eff}}=\dfrac{5\ cm}{kl_{0}A}+\dfrac{5\ cm}{kl_{0}A}$

$\Rightarrow C_{eff}=\dfrac{kl_{0}A}{(5+k5)\times 10^{-2}}$

$\Rightarrow C_{eff}=\dfrac{10\times 8.8\times 10^{-12}\times 500\times 10^{-4}}{55\times 10^{-2}}$

$\Rightarrow C_{eff}=8\ pF$

A highly conducting sheet of aluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates at distance

$\dfrac{d}{2}$ from positive plate where

$d$ is distance between plates. If the capacitance before the insertion of foil was

$10 \; \mu F$ , its value after the insertion of foil will be:

0%
$20\; \mu F$
0%
$10 \;\mu F$
0%
$5 \;\mu F$
0%
Zero

Explanation

Initial capacitor

$C=\dfrac{\epsilon_{0}A}{d}=10\mu R$
After inserting aluminium foil the effective capacitor is series combination of

$C_{1}=C_{2}=\frac{\epsilon_{0}A}{\frac{d}{2}}$

$\therefore \dfrac{1}{C_{eff}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$

$\Rightarrow C_{eff}=\dfrac{C_{1}}{2}$

$\Rightarrow C_{eff}=\dfrac{\epsilon_{0}A}{d}$

$\Rightarrow C_{eff}=10\mu F$

Charges $+q$ , $-4q$ and $+2q$ are arranged at the corners of an equilateral triangle of side $0.15\ m$ . If $q=1\ \mu C$ , their mutual potential energy is roughly:

0%
$0.4\ J$
0%
$0.5\ J$
0%
$0.6\ J$
0%
$0. 8\ J$

Explanation

We know potential energy between two points

$=\dfrac {kQ_1Q_2}{r.}$
Now P. E between A and B

$=\dfrac {k(q)(-4q)}{a}$
P. E between A and C

$=\dfrac {k(q)(+2q)}{a}$
P. E between B and C

$=\dfrac {k(-4q)(2q)}{a.}$
total P. E of arrangement

$=\dfrac {-4kq^2}{a}+\dfrac {2kq^2}{a}-\dfrac {8kq^2}{a}=\dfrac {-10kq^2}{a}$

$=\dfrac {-10\times 9\times 10^9\times 1\times 10^{-6}\times 1\times 10^{-6}}{0\cdot 15}=-0\cdot 6J.$

Calculate the electrostatic potential energy of an electron-proton system of hydrogen atom. In the first Bohr orbit of hydrogen atom, the radius of the orbit is $5.3\times 10^{-11}m\$ :

0%
$-4.35\times 10^{-18}J$
0%
$-2.175\times 10^{-18}J$
0%
$-4.35\times 10^{-19}J$
0%
$-2.175\times 10^{-19}J$

Explanation

The electrostatic P.E. of proton electron pair is

$-\dfrac{1}{4\pi \epsilon _0}\dfrac{e^2}{r}$ where

$r$ is the separation between the proton and electron and

$e$ is electron's charge.
Since the charges are opposite in sign, P.E. is negative.
Given,

$r=5.3 \times 10^{-11}\:m$
Hence,

$P.E. = -(9\times 10^9) \times \dfrac{( 1.6 \times 10^{-19})^2}{5.3 \times 10^{-11}}= -4.35 \times 10^{-18}\: J$

An oil condenser has a capacity of

$100\; \mu F$ . The oil has dielectric constantWhen the oil leaks out , its new capacity is :

0%
$200\; \mu F$
0%
$0.02\; \mu F$
0%
$50 \;\mu F$
0%
$0.5 \;\mu F$

Explanation

We know,

$C=\dfrac{\epsilon_{0}A}{d}$
When oil was present

$\epsilon_{0}$ becomes

$k\epsilon_{0}$

$\therefore$ when oil leaks out capacitors decreases by k.

$\therefore C_{leaks}=\dfrac{c}{k}$

$=\dfrac{100\mu R}{2}$

$C_{leaks}=50\mu F$

A charge of $2\ C$ is moved from a point $2\ m$ away from a charge of $1\ C$ to a point $1\ m$ away from that charge. The work done is:

0%
$10^{9}\ J$
0%
$10^{6}\ J$
0%
$9\times 10^{9}\ J$
0%
$10^{10}\ J$

Explanation

Voltage due to

$1\ C$ at a distance

$r$ is

$\dfrac {k \times 1}{r}$ .

Total change in potential in moving

$2\ C$ charge from

$2\ m$ away to

$1\ m$ away:

$V_2-V_1=\dfrac {k \times 1}{1}-\dfrac {k \times 1}{2}$
or,

$V_2-V_1=\dfrac{k \times 1}{2} = \Delta V$

So, work done

$=\Delta V Q$

$=\dfrac {k \times 1}{2}\times 2=9\times 10^9\ J$

A positive point charge q is carried from a point B to a point A in the electric field of a point charge +Q. If the permittivity of free space is $\epsilon _{0}$ , the work done in the process is given by

0%
$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a}-\frac{1}{b} \right ]$
0%
$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a}+\frac{1}{b} \right ]$
0%
$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a^{2}}-\frac{1}{b^{2}} \right ]$
0%
$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a^{2}}+\frac{1}{b^{2}} \right ]$

Explanation

Work done,

$W=q[V_2-V_1]$
Voltage at

$a, V_a=\dfrac {kQ}{a}$
Voltage at

$b, V_b=\dfrac {kQ}{b}$

$\Rightarrow W=q[\dfrac {kQ}{a}-\dfrac {kQ}{b}]$

$=\dfrac {qQ}{4\pi \varepsilon_0} \left[ \dfrac {1}{a}-\dfrac {1}{b} \right]$

A parallel plate capacitor with air between the plates is charged to a potential difference of

$500 V$ and then insulated. A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes

$75V$ . The dielectric constant of plastic is :

0%
$10/3$
0%
$5$
0%
$20/3$
0%
$10$

Explanation

$C$ be the air filled capacitance , dielectric filled capacitance becomes

$C'=KC$ where K is the dielectric constant.
Initial charge ,

$q_i=CV_i=500 C$ and final charge

$q_f=C'V_f=75KC$
As the system is isolated so charge will be constant.
Thus,

$q_i=q_f$

$\Rightarrow 500C=75 KC \Rightarrow K=\dfrac{500}{75}=20/3$

A, B, C are three points on a circle of radius $1 cm$ . These points form the corners of an equilateral triangle. A charge $2C$ is placed at the centre of the circle. The work done in carrying a charge of $0.1\mu C$ from A to B is :

0%
$Zero$
0%
$18\times 10^{11}J$
0%
$1.8\times 10^{11}J$
0%
$54\times 10^{11}J$

Explanation

Potential at a distance

$1 cm$ from centre is constant given by

$\dfrac {kQ}{(1cm)}$

$\therefore$ The circle is equipotential surface
When a charge is moved from A to B
there is no change in potential

$\therefore W=q\Delta V, \ here \Delta V=0$

$\therefore W=0$

The capacity of a parallel plate condenser is

$10 \;\mu F$ without the dielectric. Material with a dielectric constant of

$2$ is used to fill half-thickness between the plates. The new capacitance is

$\underline{\hspace{0.5in}} \;\mu F$ :

0%
$10$
0%
$20$
0%
$15$
0%
$13.33$

Explanation

Initial capacitance

$=\dfrac{\epsilon_{0}A}{d}=10\mu F$
Now effective capacitance is series combination of

$\dfrac{\epsilon_{0}A}{\dfrac{d}{2}}$ and

$\dfrac{k\epsilon_{0}A}{\dfrac{d}{2}}$

$\therefore C_{eff}=$ series of

$20\mu F$ and

$40\mu F$

$\dfrac{1}{C_{eff}}=\dfrac{1}{20}+\dfrac{1}{40}$

$C_{eff}=13.33 \mu F$

The plates of a parallel plate capacitor are charged up to

$200$ volts. A dielectric slab of thickness 4mm is inserted between the plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by

$3.2mm$ . The dielectric constant of a dielectric slab is :

0%
$1$
0%
$4$
0%
$5$
0%
$6$

Explanation

Initial voltage

$=200 V$

$q=C.V$

$q=\dfrac{\epsilon_0.A}{d}V$ (1)
When dielectric is inserted

$q$ remains constant and V decreases by k.

$q=\dfrac{\epsilon_0A}{d'+t-\dfrac{t}{k}}V$ (2)

From equation 1 and 2 we have,

$d=d'+t-\dfrac{t}{k}$

$d-d'=1.6mm=2-\dfrac{2}{k}$

solving,

$k=5$

The capacitance of a capacitor becomes $\dfrac{7}{6}$ times its original value if a dielectric slab of thickness, $t=\dfrac{2}{3}d$ is introduced in between the plates. d is the separation between the plates. The dielectric constant of the dielectric slab is :

0%
$\dfrac{14}{11}$
0%
$\dfrac{11}{14}$
0%
$\dfrac{7}{11}$
0%
$\dfrac{11}{7}$

Explanation

Now

$C_{eff}$ is series combination of

$\dfrac{l_{0}A}{x}, \dfrac{l_{0}A}{d-t-x}\ and\ \dfrac{kl_{0}A}{t}$

$\dfrac{1}{C_{eff}}=\dfrac{x}{l_{0}A}+\dfrac{d-t-x}{l_{0}A}+\dfrac{t}{kl_{0}A}$

$\Rightarrow C_{eff}=\dfrac{kl_{0}A}{(d-t)k+t}$

$\Rightarrow$ given

$C_{eff}=\dfrac{7}{6}\cdot \dfrac{l_{0}A}{d}$

$\dfrac{7}{6}\cdot \dfrac{l_{0}A}{d}=\dfrac{kl_{0}A}{(d-\dfrac{2}{3}d)k+\dfrac{2}{3}d}$

$\Rightarrow \dfrac{7}{6}=\dfrac{3k}{(\dfrac{1}{3})k+\dfrac{2}{3}}$

$\Rightarrow 7k+14=18k$

$\Rightarrow k=\dfrac{14}{11}$

The potential difference across $3 \mu F$ condenser is :

0%
$40 Volt$
0%
$60 Volt$
0%
$80 Volt$
0%
$120 Volt$

Explanation

As both the capacitor are in series the charge flowing through it is same,

$Q_1=Q_2$
Now if we consider potential across

$3\mu F$ capacitor as V then potential across

$6\mu F$ capacitor is given as

$120-V$ ,

$(120-V)6\mu F=V(3\mu F)$

$720 \mu F =V(9\mu C)$

$V = 80 V$

The equivalent capacitance of three capacitors of capacitance $C_{1}, C_{2}$ and $C_{3}$ connected in parallel is 12 units and the product $C_{1}C_{2}C_{3}=48$ . When the capacitors $C_{1}$ and $C_{2}$ are connected in parallel the equivalent capacitance is 6 units. Then the capacitance are :

0%
$1.5, \ 2.5, \ 8$
0%
$2,\ 3, \ 7$
0%
$4, \ 2, \ 6$
0%
$1, \ 5, \ 6$

Explanation

$C_{1},\ C_{2}$ and

$C_{3}$ in parallel

$C_{eff}=C_{1}+C_{2}+C_{3}$

$12=C_{1}+C_{2}+C_{3}$ ------------ (1)

$C_{1}C_{2}C_{3}=48$ ------------ (2)

$C_{1}$ and

$C_{1}$ are connected in parallel

$C_{eff}=C_{1}+C_{2}$

$C_{1} +C_{2} = 6$ --------(3)
from(1) and (3),

$C_{3} = 6$
Hence equation (3) becomes

$C_{1}C_{2}=8$ ------(4)
also from (2)

$C_{1}=6-C_{2}$

$(6-C_{2})C_{2} = 8$

$C_{2}^2-6C_{2}+8=0$
solving this we get,
We get

$C_{1}=4$ ,

$C_{2}=2$ ,

$C_{3}=6$

Two condensers of capacitance $4\mu F$ and $5\mu F$ are joined in series. If the potential difference across $5\mu F$ is $10V$ , then the potential difference across $4\mu F$ condenser is :

0%
$22.5V$
0%
$10V$
0%
$12.5V$
0%
$25V$

Explanation

When two capacitors are in series

$Q$ present on then is same

$\therefore\ Q\ on \ 5\mu F$ is

$Q=5\times10^{-6}\times 10$

$Q=5\times 10^{-5}$

$\therefore Q$ on

$4\mu F$ is

$5\times 10^{-5}$

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

$V=\dfrac{5\times 10^{-5}}{4\times 10^{-6}}$

$V=12.5\ V$

In the capacitor of capacitance $20\ \mu F$ , the distance between plates is $2\ mm$ . If a material of dielectric constant $2$ is inserted between the plates, then the capacitance of the system is :

0%
$20 \mu F$
0%
$30 \mu F$
0%
$22 \times 5 \mu F$
0%
$40 \mu F$

Explanation

We know

$C=\dfrac{\epsilon_{0}A}{d}$
When k is added

$C_{1}=\dfrac{k\epsilon_{0}A}{d}$

$\Rightarrow C_{1}=2\times 20$

$\Rightarrow C_{1}=40 \mu F$

A charge

$3$ coulomb experiences a force

$3000\ N$ when placed in a uniform electric field. The potential difference between two points seperated by a distance of

$1\ cm$ along the field lines is :

0%
$10\ V$
0%
$90\ V$
0%
$1000\ V$
0%
$9000\ V$

Explanation

Electric force $F=qE$

given, $F=3000 N, q=3 C$

$\therefore E=\dfrac{F}{q}=\dfrac{3000}{3}=1000 V/m$

Potential difference, $V={E}{d}$ where $d=0.01 m$

$\therefore V=1000\times 0.01=10 V$

Ans:(A)

Two metal plates are separated by a distance $d$ in a parallel plate condenser. A metal plate of thickness $t$ and of the same area is inserted between the condenser plates. The value of capacitance increases by a factor of :

0%
$\dfrac{d-t}{d}$
0%
$\left ( 2-\dfrac{t}{d} \right )$
0%
$\left ( t-\dfrac{t}{d} \right )$
0%
$\dfrac{1}{\left ( 1-\dfrac{t}{d} \right )}$

Explanation

Hint:

Capacitance of a parallel plate capacitor with distance $d$ between the plates is given by,

$C=\dfrac{A{{E}_{o}}k}{d}$

Where,

$A$ is area of plates,

${{E}_{o}}$ is permittivity of free space,

$k$ is dielectric constant of medium filled between plates.

Capacitance of a parallel plate capacitor with distance $d$ between the plates, with a dielectric slab of thickness $t$ and dielectric constant $k$ between the plates is given by,

$C=\dfrac{A{{E}_{o}}}{d-t\left( 1-\dfrac{1}{k} \right)}$

Note that for metal, value of $k$ is infinite.

Step 1: Given.

A parallel plate capacitor with distance d between plates of area A. Capacitance is given by,

$C=\dfrac{A{{E}_{o}}}{d}$

Step 2: Calculate new capacitance in terms of old capacitance.

A metal plate of thickness t is inserted between them. New capacitance is:

$C'=\dfrac{A{{E}_{o}}}{d-t\left( 1-\dfrac{1}{k} \right)}$

$\Rightarrow C'=\dfrac{A{{E}_{o}}}{d-t}$ (Since, k is infinite for metal)

$\Rightarrow C'=\dfrac{A{{E}_{o}}}{d(1-\dfrac{t}{d})}$

$\Rightarrow C'=\dfrac{C}{(1-\dfrac{t}{d})}$

$\Rightarrow C'=\dfrac{1}{(1-\dfrac{t}{d})}C$
Thus value of capacitance increases by a factor $\dfrac{1}{(1-\dfrac{t}{d})}$

Option D is correct.

Three capacitors $2\mu F, 3\mu F$ and $5\mu F$ are connected in parallel. The capacitance of the combination:

0%
$\dfrac{30}{31}\mu F$
0%
$\dfrac{31}{30}\mu F$
0%
$10 \mu F$
0%
$2.5 \mu F$

Explanation

When connected in parallel ,

$C_{eff}=C_{1}+C_{2}+C_{3}$

$C_{eff}=2+3+5$

$C_{eff}=10\mu F$

The equivalent capacity between the points X and Y in the circuit with $C=1\ \mu F$ is:

0%
$2\ \mu F$
0%
$3\ \mu F$
0%
$1\ \mu F$
0%
$0.5\ \mu F$

Explanation

The second capacitor, between the two conducting wires, is short circuited.

Thus no charge flows through it.

$\therefore C_{eff}=C_{1}+C_{2}$

$=2C=2\ \mu F$

A capacitor of $10 \mu F$ capacitance is charged by a $12 V$ battery. Now the space between the plates of capacitors is filled with a dielectric of dielectric constant $K = 3$ and again it is charged. The magnitude of the charge is :

0%
$120 \mu C$
0%
$240 \mu C$
0%
$360 \mu C$
0%
$480 \mu C$

Explanation

$\textbf{Step 1: Initial charge [Refer Fig. 1]}$

$Q_1 = C_1 V_1 = 10\ \mu F \times 12\ V$

$= 120\ \mu C$

$\textbf{Step 2: Final capacitance after insertion of dielectric [Refer Fig. 2]}$

$C_{2} = \dfrac{k \epsilon_0 A}{d} = kC_{1} = 3 \times 10\ \mu F$

$C_{2} = 30\ \mu F$

$\textbf{Step 3: Final charge [Refer Fig. 3]}$

As the battery remains connected, voltage across capacitor remains same.

i.e.,

$V_{2} = V_{1} = 12\ V$

$Q_{2} = C_{2} V_{2}$

$Q_{2} = 30\ \mu F \times 12\ V$

$= 360\ \mu C$

Hence, option C is correct.

Three capacitors $2\mu F, 3\mu F$ and $6\mu F$ are connected in series. The effective capacitance of the combination is:

0%
$11\ \mu F$
0%
$1\ \mu F$
0%
$1.2\ \mu F$
0%
$2\ \mu F$

Explanation

When capacitors are in series

$\dfrac{1}{C_{eff}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}$

$=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}$

$\Rightarrow C_{eff}=1\ \mu F$

A neutral hydrogen molecule has two protons and two electrons. If one of the electrons is removed we get a hydrogen molecular ion $\left ( H_{2}^{+} \right )$ . In the ground state of $H_{2}^{+}$ the two protons are separated by roughly $1.5$ $\ A^o$ and the electrons is roughly $1$ $A^o$ from each proton. The potential energy of the system is :

0%
$-38.4\ eV$
0%
$-19.2\ eV$
0%
$-9.6\ eV$
0%
$zero$

Explanation

The energy of the system is given by,

$ke^2/(1.5 \,angstrom) + 2( k(-e^2)/(1 \, angstrom) )$

$= 9\times 10^9\times 10^{10} \times ( 2e^2/3 - 2e^2 )$

$= 9\times 10^{19} \times (-4/3) e \,\,eV = -19.2 \,\,eV$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 2 - 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 2 - MCQExams.com

0:0:4

Practice Class 12 Medical Physics Quiz Questions and Answers

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