MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 3

Two equal point charges are fixed at $x = -a$ and $x= +a$ on the x- axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to :

0%
$x$
0%
$x^{2}$
0%
$x^{3}$
0%
$1 / x$

Explanation

Potential of the charge when displaced by a distance x from origin

$=\dfrac{kQq}{a-x}+\dfrac{kQq}{a+x}=\dfrac{kQq(2a)}{a^2-x^2}$

Initial potential

$=\dfrac{2kQq}{a}$

Hence the change in potential

$=\dfrac{kQq(2a)}{a^2-x^2}-\dfrac{2kQq}{a}=\dfrac{2kQqx^2}{(a^2-x^2)a}$

For small values of

$x, a^2-x^2\approx a^2$

Hence the change is

$\dfrac{2kQqx^2}{a^3}$ .

A parallel plate condenser has initially air medium between the plates. If a slab of dielectric constant 5 having thickness half the difference of separation between the plates is introduced, the percentage increase in its capacity is :

0%
33.3%
0%
66.7%
0%
50%
0%
75%

Explanation

$t=\dfrac{d}{2}, K= 25$

$C=\dfrac { { \varepsilon }_{ o }A }{ \left[ \left( d-t \right) +\dfrac { t }{ k } \right] }$

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

Initially capacitance is C

${ C }^{ 1 }=\dfrac { { \varepsilon }_{ o }A }{ \left[ d-\dfrac { d }{ 2 } +\dfrac { d }{ 2\left( 5 \right) } \right] }$

${ C }^{ 1 }=\dfrac { { \varepsilon }_{ o }A }{ \dfrac { d }{ 2 } \left[ 1+\dfrac { 1 }{ 5 } \right] }$

${ C }^{ 1 }=\dfrac { { \varepsilon }_{ o }A }{ d\left[ 1+\dfrac { 1 }{ 5 } \right] }$

Therefore % increase =

$\left[ \dfrac { 10 }{ 6 } -1 \right] \times 100=66.7\%$

Two identical metal plates separated by a distance

$d$ forms parallel plate capacitor of capacity

$C$ . A metal sheet of thickness

$d/2$ and same dimensions is inserted between the plates, so that the air gap is separated into two equal parts. The new capacity of the system will be:

0%
$\displaystyle \dfrac{\varepsilon _{0}A}{2d}$
0%
$\displaystyle \frac{\varepsilon _{0}A}{4d}$
0%
$\displaystyle \frac{2\epsilon_{0}A}{d}$
0%
$\displaystyle \frac{\epsilon_{0}A}{d}$

Explanation

The final capacitance in series combination of

$\dfrac{ \varepsilon _0 \ A}{\dfrac{d}{4}}$ and

$\dfrac{ \varepsilon _0\ A}{\dfrac{d}{4}}$ is

$\dfrac{1}{C} = \dfrac{2d}{4\epsilon_0 A }$
Hence ,

$C = \dfrac{2\epsilon_0 A }{d}$

Three capacitors $3 \mu F, 10 \mu F$ and $15 \mu F$ are connected in series to a voltage source of 100V. The charge on $15 \mu F$ is :

0%
$22 \mu C$
0%
$100 \mu C$
0%
$2800 \mu C$
0%
$200 \mu C$

Explanation

$\dfrac{1}{C_{ef}}=\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{15}$

$\dfrac{1}{C_{ef}}=\dfrac{1}{3}+ \dfrac{5}{30}=\dfrac{3}{6}$

$\therefore C_{eff} = 2$

$\therefore$ Charge of

$C_{eff}$ is

$q=C_{eff}\times V$

$q = (2) (100)$

$q = 200 \mu C$
In series all the capacitors have same charge

$\therefore charge \ on \ 15 \mu \ F \ is \ 200 \mu C$

A condenser is charged to a potential difference of $120 \; V$ . It's energy is $1 \times 10^{-5}\; J$ . If battery is there and the space between plates is filled up with a dielectric medium $\left ( \varepsilon _{r}=5\right )$ . Its new energy is

0%
$10^{-5}J$
0%
$2 \times 10^{-5}J$
0%
$3 \times 10^{-5}J$
0%
$5 \times 10^{-5}J$

Explanation

Given energy =

$1\times 10^{-5}\$
When battery is connected,

$V$ across capacitor will be constant.
When dielectric is filled,

$C$ increases by

$K$

$\therefore New\ energy\ = \dfrac{1}{2}(KC) V\ ^{2}$

$=K\dfrac{1}{2} C V\ ^2 ( Given: \dfrac{1}{2} C V^2 = 1 \times 10^{-5} \ J)$

$= K (1 \times 10^{-5})\ J$

$= 5 \times (1 \times 10^{-5}) \ J$

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

A condenser of

$1\mu F$ is charged to a potential of

$1000 V$ . If a dielectric slab of dielectric constant

$5$ is introduced between the plates of the condenser after disconnecting the battery, the loss in the energy of the condenser is :

0%
0.1 J
0%
2.5 J
0%
0.4 J
0%
5 J

Explanation

After disconnecting battery charge present on capacitor becomes constant

$Q= 1\times 10^{-6}\times 1000$

$Q= 10^{-3}C$
When dielectric is added C becomes KC,

$\therefore$ V decreases to

$\dfrac{V}{K}$

$\therefore$ loss in energy

$= \dfrac{1}{2}CV^2-\dfrac{1}{2}(KC)(\dfrac{V}{K})^2$

$= \dfrac{1}{2}CV^2-\dfrac{1}{2}K\ C\dfrac{V^2}{K^2}$

$= \dfrac{1}{2}C\ V^2(1-\dfrac{1}{K})$

$= \dfrac{1}{2}\times 10^{-6}\times 10^6(1-\dfrac{1}{5})$

$= \dfrac{1}{2}(\dfrac{4}{5})$

$= 0.4J$

The area of the positive plate is $125\;cm^{2}$ and the area of the negative plate is $100\;cm^{2}$ , They are parallel to each other and are separated by 0.5 cm . The capacity of a condenser with air as dielectric is :

$\left ( \varepsilon _{0}=8.9\times 10^{-12}\;C^{2}N^{-1}M^{-2} \right )$

0%
$22.25\; pF$
0%
$20.02 \;pF$
0%
$17.8\; \mu F$
0%
$17.8 \;pF$

Explanation

when area of plate are different we consider the area which is smaller to calculate capacitor,

$\therefore c=\dfrac{\epsilon_\circ{}\ A}{d}$

$\Rightarrow c=\dfrac{8.9\times 10^{-12}\times 100\times 10^{-4}}{0.5\times 10^{-2}}$

$\Rightarrow c=17.8\ pF$

Two identical metal plates, separated by a distance d form a parallel plate capacitor. A metal sheet of thickness $\dfrac{d}{2}$ of the same area as that of either plate, is inserted between the plates. The ratio of the capacitance's after the insertion of the sheet to that before insertion is:

0%
$\sqrt{2}:1$
0%
2 :1
0%
1:1
0%
$1:\sqrt{2}$

Explanation

The effective capacitance of above capacitance is series combination of $\displaystyle \frac{ {4\epsilon}_{0} A}{d}$ and $\dfrac{{4\epsilon}_{0}A }{d}$

$\Rightarrow \dfrac{1}{C_{q}}=\dfrac{d}{4 {\epsilon}_{0} A}+\dfrac{d}{4 {\epsilon}_{0}A}$

$\Rightarrow C_{eff}=\dfrac{2 {\epsilon}_{0}A}{d}$

$\dfrac{C_{2}}{C_{1}}=\dfrac{\dfrac{2 {\epsilon}_{0}A}{d}}{\dfrac{{\epsilon}_{0}A}{d}}=2:1$

When a dielectric slab of thickness $4 \;cm$ is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by $3 \; cm$ to restore the capacity to it's original value. The dielectric constant of the slab is

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

Explanation

$\textbf{Step 1 - Writing capacitance equation}$

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

$....(1)$

When dielectric is added

$C' = \dfrac {\varepsilon_{0}A}{d - t + \dfrac {t}{k}}$

given that distance has to be increased by

$3\ cm$ to restore original capacitance

$\Rightarrow C' = C = \dfrac {\varepsilon_{0}A}{d + 3 - t + \dfrac {t}{k}}$

$....(2)$

$\textbf{Step 2 - Calculating dielectric constant}$

Equating equations

$(1)$ and

$(2)$

$\dfrac {\varepsilon_{0}A}{d} = \dfrac {\varepsilon_{0}A}{d + 3 - t + \dfrac {t}{k}}$

$\Rightarrow d = d + 3 - 4 + \dfrac {4}{k}$

$(\because t = 4\ cm)$

$\Rightarrow k = 4$

Hence, the dielectric constant of the slab is

$4$ .

A electric charge

${ 10 }^{ -8 }C$ is placed at the point

$(4m,7m,2m)$ . At the point

$(1m,3m,2m)$ the electric

0%
potential will be $18V$
0%
field has no $Y$ -component
0%
field will be along $Z$ -axis
0%
potential will be $1.8V$

Explanation

Electric potential $,\ V=\dfrac{1}{4\pi \epsilon_0}\dfrac{q}{r}=\dfrac{1}{4\pi \epsilon_0}\dfrac{q}{\sqrt{(4-1)^2+(7-3)^2+(2-2)^2}}=\dfrac{10^{-8}}{4\pi \times 8.854\times 10^{-12}\times 5}=18 volt$

1 V equals :

0%
1 J
0%
1 JC $^{-1}$
0%
1 CJ $^{-1}$
0%
1 JC

Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. $Q_{1}$ and $Q_{2}$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $q_{1}$ and $q_{2}$ . Then :

0%
$\frac{q_{1}}{Q_{1}}=\frac{K+1}{K}$
0%
$\frac{q_{2}}{Q_{2}}=\frac{K+1}{2}$
0%
$\frac{q_{2}}{Q_{2}}=\frac{K+1}{2K}$
0%
$\frac{q_{2}}{Q_{2}}=\frac{K}{2}$

Explanation

In initial case,

$C_{net} = ( C \times K C)/(C + K C) = CK (1+K)$
In final case,

$C_{net} = C^2/2C = C/2$
Now,

$Q_{initial}/Q_{final} = C_{initial}/C_{final} = 2K/( 1+K)$

Two capacitors of capacities 3 $\mu$ F and 6 $\mu F$ are connected in series and connected to 120V. The potential differences across 3 $\mu$ F is $V_{0}$ and the charge here is $q_{0}$ . We have :

A) $q_{0}=40\mu C$ B) $V_{0}=60V$

C) $V_{0}=80V$ D) $q_{0}=240\mu C$

0%
A, C are correct
0%
A, B are correct
0%
B, D are correct
0%
C, D are correct

Explanation

When capacitor are connected is series charge present on capacitor is same.
Now,

$q_0 =CV=240\times 10^{-6}$
We know,

$V_0=\dfrac{q_0}{C}$

$V_0=\dfrac{{240}\times 10^{-6}}{3\times 10^{-6}}$

$V_0=80V$

A potential difference of $300$ volts is applied to a combination of $2.0\mu$ F and $8.0\mu$ F capacitors connected in series. The charge on the $2.0\mu$ F capacitor is :

0%
$2.4\times 10^{-4}coulomb$
0%
$4.8\times 10^{-4}coulomb$
0%
$7.2\times 10^{-4}coulomb$
0%
$9.6\times 10^{-4}coulomb$

Explanation

We know when capacitor are connected in series charge present on them must be equal.
In seris

$\dfrac{1}{C_{s}}= \dfrac{1}{C_1}+\dfrac{1}{C_2}$

$\dfrac{1}{C_{s}}= \dfrac{1}{2}+\dfrac{1}{8}$

$\dfrac{1}{C_{s}}= \dfrac{5.}{8}$

$C_{s}= \dfrac{8}{5} \mu F$

$Q=CV$

$Q= \dfrac{8}{5}\times 10^{-6}\times 300$

$Q= 4.8\times 10^{-4}C.$

The capacitors of three capacities are in the ratio 1 : 2 :Their equivalent capacity when connected in parallel is $\frac{60}{11}\mu F$ more than that when connected in series. The individual capacities are :

0%
4, 6, 7
0%
1, 2, 3
0%
2, 3, 4
0%
1, 3, 6

Explanation

Let the capacitance be,
C, 2C and 3C .

in series

$\dfrac{1}{C_{eff}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$

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

$\Rightarrow \dfrac{1}{C_{eff}}=\dfrac{6+3+2}{6 C}$

$\Rightarrow c_{eff}=\dfrac{6 C}{11}$

in parallel

$C_{eff}=C_1+C_2+C_3$

$\Rightarrow C_{eff}=C+2C+3C$

$=6C$
given parallel connection

$=\dfrac{60}{11}\mu \ F$ more than the series connection

$\Rightarrow 6C=\dfrac{60}{11}\mu \ F+\dfrac{6\ C}{11}$

$\Rightarrow \dfrac{60\ C}{11}=\dfrac{60}{11}\mu \ F$

$\Rightarrow C=1\mu F ; 2C=2 \mu F ; 3C=3\mu F$

A parallel plate capacitor with plates separated by air acquires 1 $\mu$ C of charge when connected to a battery of 500V. The plates still connected to the battery are then immersed in benzene ( $k=2.25$ ). Then a charge flows from the battery is :

0%
$1.25 \mu$ C
0%
$2.28 \mu$ C
0%
$1/4 \mu$ C
0%
$4.56 \mu$ C

Explanation

Initial charge

$Q_1=1\mu c$

$V_1=500V.$
As the plates are still connected voltage is constant.

$\therefore$ when plates are dipped in benzene capacitance increases by

$k=(\dfrac {k\varepsilon_0 A}{d})$

$\therefore$ As voltage is constant Q also increases by k.

$\therefore Q_2=kQ_1$
charge flown from battery

$=kQ_1-Q_1$

$1\mu c(2\cdot 25-1)$

$1\times 10^{-6}(1\cdot 25)$

$=1\cdot 25\mu c.$

A capacitor of capacitance $C$ has charge $Q$ and stored energy $W$ . If the charge is increased to $2Q$ , the stored energy would be:

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

Explanation

We know that the energy stored in a capacitor is given by:

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

Now, if the charge

$Q$ increases to

$2Q$ ,

then the change in energy will be given by:

$W'=\dfrac{(2Q)^2}{2C}$

$\Rightarrow\dfrac{4Q^{2}}{2C}$

$=4W$

Two identical parallel plate capacitors are joined in series to

$100\;V$ battery. Now a dielectric with

$K=4$ is introduced between the plates of second capacitor. The potential difference on capacitors now becomes

0%
$60\; V, 40 \; V$
0%
$70\; V, 30\; V$
0%
$75\; V, 25\; V$
0%
$80 \; V, 20 \; V$

Explanation

Introducing a dielectric increases the capacitance by

$K$ .
Now, charge on both the capacitors is same

$\Rightarrow CV_1 = 4CV_2$
Also,

$V_1 + V_2 = 100\; V \Rightarrow 5V_2 = 100 \Rightarrow V_2 = 20 \; V$

$V_1=100-20 \;V=80 \;V$

Potential across capacitors

$=80V,20V$

The distance between the plates of a condenser is reduced to $\frac{1}{4}th$ and the space between the plates is filled up by a medium of dielectric constant K(2.8). The capacity is increased by :

0%
5.6times
0%
11.2times
0%
22.4 times
0%
44.8 times

Explanation

Initially,

$\ C=\dfrac{\epsilon_0\ A}{d}$

$\Rightarrow$ Distance is reduced by

$\frac{1}{4}$ and K is added

$\therefore C_{new}=\dfrac{k\ \epsilon_0\ A}{(^{d}/_{a})}$

$\Rightarrow C_{new}=2.8\times \dfrac{\epsilon_0\ A}{d}\times 4$

$\Rightarrow C_{new}=11.2\times C.$

The capacity of a parallel plate condenser with air medium is C . If half of the space between the plates is filled with a slab of dielectric constant K as shown in the figure, then the capacity becomes :

0%
$\frac{K}{2}C$
0%
2KC
0%
$\frac{\left ( K+1 \right )C}{2}$
0%
(K+1)C

Explanation

Capacity of a half parallel plate capacitor with air medium

$= \dfrac{C}{2}$

Capacity of a half parallel plate capacitor with slab of dielectric constant K

$=\dfrac{KC}{2}$

$\therefore$ Effective capacitance=capacity=

$C_{eff}=\dfrac{KC}{2}+\dfrac{C}{2}=\dfrac{(K+1)C}{2}$

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

0%
6E, 6C
0%
E, C
0%
$\dfrac {E}{6},6C$
0%
E, 6C

Explanation

Let C be charged to v

$\Rightarrow E=\dfrac{1}{2}\times C\times V^{2}$
Now when dielectric is added.
C increases by kC and as battery is disconnected

$\upsilon$ is constant.

$\therefore$ V decreases by k.

$\therefore C_{1}=kC; V_{1}=\dfrac{V}{k}$

$\Rightarrow E_{1}=\dfrac{1}{2}c_{1}v_{1}^{2}$

$E_{1}=\dfrac{1}{2}kc\times \dfrac{v}{k}\times \dfrac{v}{k}$

$E_{1}=\dfrac{1}{2}c\dfrac{v^{2}}{6}$

$c_{1}=\dfrac{1}{6}\cdot E$

$\therefore E_{1}=\dfrac{E}{6}$
and

$C_{1}=6C$

If a proton and an electron are accelerated through the same potential difference then:

0%
both the proton and electron have same K.E
0%
both the proton and electron have same momentum
0%
both the proton and electron have same velocity
0%
both the proton and electron have same temperature

The work done in placing a charge of

$8\times 10^{-18}$ coulomb on a condenser of capacity

$100$ micro-farad is:

0%
$16\times 10^{-32}$ joule
0%
$3.2\times 10^{-26}$ joule
0%
$4\times 10^{-10}$ joule
0%
$32\times 10^{-32}$ joule

Explanation

Here the work done

$=$ energy stored in capacitor

$=\dfrac{Q^2}{2C}$
or,

$W=\dfrac{(8\times 10^{-18})^2}{2\times 100\times 10^{-6}}=32\times 10^{-32}\ J$

A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric filed and energy associated with this capacitor dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities are now given by $Q$ , $V$ , $E$ and $U$ are related to the previous one as:

0%
$Q > Q_{0}$
0%
$V > V_{0}$
0%
$E > E_{0}$
0%
$U > U_{0}$

Explanation

$Q_0=C_0 V_0$

$U_0=\dfrac {1}{2}C_0 V_{)}^{2}$

$E_0=\dfrac {V_0}{(d)}$ (d is distance between plates).
when dielectric is added.

$c=kC_0$
As battery is still connected

$U_0$ remains same.

$V=U_0$

$U=\dfrac {1}{2}(kC_0)V_{0}^{2}$

$U-k\dfrac {1}{2} C_0 V_{0}^{2}$

$U=k V_0$

$Q=k C_0\cdot V_0.$

$\therefore Q=k Q_0$

$E=\dfrac {U}{d}$

$E=\dfrac {U_0}{d}$

$E=E_0.$

A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature of the block is raised by

$\Delta$ T, the potential difference V across the capacitance is :

0%
$\sqrt{\dfrac{2mC\Delta T}{s}}$
0%
$\dfrac{mC\Delta T}{s}$
0%
$\dfrac{ms\Delta T}{C}$
0%
$\sqrt{\dfrac{2ms\Delta T}{C}}$

Explanation

The energy of capacitor is used to raise the temperature of the body.

$1/2\times CV^2 = ms\Delta T \Rightarrow V = \sqrt{ \dfrac{2ms\Delta T}{C}}$

Between the plates of a parallel plate capacitor of capacity C, two parallel plates of the same material and area same as the plate of the original capacitor, are placed. If the thickness of these plates is equal to

$\left\{\dfrac {1}{5}\right\}^{th}$ of the distance between the plates of the original capacitor, then the capacity of the new capacitor is :

0%
$\dfrac {5}{3}C$
0%
$\dfrac {3}{5}C$
0%
$\dfrac {3}{10}C$
0%
$\left (\dfrac {10}{3} \right )C$

Explanation

Initial capacitance

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

Now assume the plates are arranged as follows.

The above arrangement acts as three capacitors kept in series.

$C_1=\dfrac {\varepsilon_0 A}{x_1}, C_2=\dfrac {\varepsilon_0 A}{x_2}, C_3=\dfrac {\varepsilon_0 A}{x_3}$

$\dfrac {1}{C_{eff}}=\dfrac {1}{C_1}+\dfrac {1}{C_2}+\dfrac {1}{C_3}$

$\Rightarrow \dfrac {1}{C_{eff}}=\dfrac {x_1+x_2+x_3}{\varepsilon_0 A}$

$\Rightarrow C_{eff}=\dfrac {\varepsilon_0 A}{x_1+x_2+x_3}$

$x_1+x_2+x_3=\dfrac {3d}{5}$

$\therefore C_{eff}=\dfrac {5\varepsilon_0 A}{3d}=\dfrac {5C}{3.}$

The capacitance of a parallel plate condenser is $C_{1}$ (fig. a). A dielectric of dielectric constant ‘K’ is inserted as shown in figure ‘b’ and ‘c’. If $C_{2}$ and $C_{3}$ are the capacitances in figures ‘b’ and ‘c’ then :

0%
Both C $_{2}$ and C $_{3}$ >C $_{1}$
0%
C $_{3}$ >C $_{1}$ and C $_{2}$ >C $_{1}$
0%
Both C $_{2}$ and C $_{3}$ < C $_{1}$
0%
C $_{1}$ =C $_{2}$ =C $_{3}$

Explanation

We know that capacitance is given by

$C_1=\dfrac {\varepsilon_0 A}{d.}$
b) In

$C_2\dfrac {k \varepsilon_0 A 2}{d} and \dfrac {\varepsilon_o A 2}{d}$ are in series

$\Rightarrow \dfrac {1}{C_2}=\dfrac {d}{k\varepsilon A 2}+\dfrac {d}{2\varepsilon_0 A}=\dfrac {d}{2\varepsilon_0 A}(\dfrac {1}{k}+1)$

$\Rightarrow C_2=\dfrac {\varepsilon_0 A}{d}(\dfrac {2k}{1+k})$ we know k>1

$\therefore \dfrac {2k}{1+k}>1.$

$\therefore C_2>C_1.$

c) in

$C_3\dfrac {k\varepsilon_0 A}{2d} and \dfrac {k\varepsilon_o A}{d}$ are in parallel

$\therefore C_3=\dfrac {\varepsilon_0 A}{d}(\dfrac {k}{2}+1)$

$\therefore C_3>C_1.$

The energy stored in the capacitor in steady state is :

0%
$12\mu$ J
0%
$24\mu$ J
0%
$36 \mu$ J
0%
$48\mu$ J

Explanation

In steady state, the current flowing through capacitor branch is zero.

$I = \dfrac{(8-3)}{4+1} = 1A$
Potential of point

$P = 8-4 = 4V$
Voltage across capacitor

$= 4V$
Energy stored in capacitor

$= \dfrac{1}{2} CV^{2}$

$= \dfrac{1}{2} \times 3 \times 10^{-6} \times 16$

$= 24 \mu J$

Two short dipoles are situated parallel to each other separated by a distance x. The force of interaction is

0%
A force of attraction of $\frac{2Kp_1p_2}{x^4}$
0%
A force of repulsion of $\frac{2Kp_1p_2}{x^4}$
0%
A force of attraction of $\frac{3Kp_1p_2}{x^4}$
0%
A force of repulsion of $\frac{3Kp_1p_2}{x^4}$

Explanation

Electrostatic field due to dipole 1

$E=\frac{kp_1}{x^3}$
now second dipole is mode of two charges say one is at a distance x of other at (x + d),

$d\rightarrow$ length of dipole

$\because$ force of charge 1,

$F_1=\frac{kp_1q}{x^3}$
force on charge 2,

$F_2=\frac{-kp_1q}{(x+d)^3}$
both will be in opposite direction, as one is positive charge and other is negative net force

$f = f_1+f_2$
=

$kp_1q\left ( \frac{1}{x^3}-\frac{1}{(x+d)^3} \right )$
=

$kp_1q\left ( \frac{x^3+3x^2d+3d^2x+d^3-x^3}{x^3(x+d)^3} \right )$
=

$k{ p }_{ 1 }q\frac { \left( 3\left( \frac { d }{ x } \right) +3\left( \frac { d }{ x } \right) ^{ 2 }+\left( \frac { d }{ x } \right) ^{ s } \right) }{ (d+x)^{ 3 } }$
as d < < x ignoring higher power of d/x and d as compound to r
=

$\frac{kp_1q3\left ( \frac{d}{x} \right )}{x^3}$
=

$\frac{3kp_1(qd)}{x^4}$
=

$\frac{3kp_1p_2}{x^4}$

The electric potential at a point in free space due to a charge Q coulomb is

$Q\times 10^{11}$ volts. The electric field at that point is -

0%
$4\pi \epsilon _{0}\times 10^{20}volt/m$
0%
$12\pi \epsilon _{0}\times 10^{22}volt/m$
0%
$4\pi \epsilon _{0}\times 10^{22}volt/m$
0%
$12\pi \epsilon _{0}\times 10^{20}volt/m$

Explanation

Since

$V=\frac{Q}{4\pi \varepsilon _{0}r}$ and

$E=\frac{Q}{4\pi \varepsilon _{0}r^{2}}$ ;
Given

$V = Q \times 10^{11} = Q 4\pi \varepsilon _{0}r$

$\Rightarrow r = \frac{1}{4\pi \varepsilon _{0}\times 10^{11}}$
i.e.,

$E=\frac{V}{r}$

$\Rightarrow E=\frac{QV}{r}=Q \times 10^{11} \times 4\pi \varepsilon _{0}\times 10^{11}$

$=4\pi \varepsilon _{0}Q\times10^{22}volt \:m^{-1}$

Complete the following statements with an appropriate word /term be filled in the blank space(s).

The equivalent capacitance C for the series combination of three capacitance

$C_1,C_2$ and

$C_3$ is given by

$\cfrac{1}{C} =$ ..............

0%
$C_1+C_2+C_3$
0%
$\left ( \cfrac{1}{C_{1}+C_{2}+C_{3}} \right )$
0%
$\left ( \cfrac{1}{\cfrac{1}{C_{1}}+\cfrac{1}{C_{2}}+\cfrac{1}{C_{3}}}\right )$
0%
$\left ( \cfrac{1}{C_{1}}+\cfrac{1}{C_{2}}+\cfrac{1}{C_{3}}\right )$

Explanation

When in series, the reciprocal of the net capacitance is equal to the sum of reciprocal of individual capacitances.

The equivalent capacitance of the pair of capacitors is

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

$\cfrac{1}{C} = \cfrac{V}{Q} = \cfrac{(v_1 + v_2+ v_3)}{ Q }=\cfrac{v_1}{Q} + \cfrac{v_2}{Q}+ \cfrac{v_3}{Q} = \cfrac{1}{C_1} + \cfrac{1}{C_2}+\cfrac{1}{C_3}$

Point charges

$q_1 = +1 \: \mu C$ and

$q_2$ whose magnitude is

$64=27\:\mu C$ are fixed 5 m apart along a vertical line with

$q_1$ being at lower position. These two charges together are able to hold an oil drop of mass

$1 \mu g$ and charge Q stationary when it is 3 m away from

$q_1$ and 4 m away from

$q_2$ . The sign of the charge

$q_2$ and the value of Q are respectively:

0%
$q_2 \: is \: positive, Q = 6.25 \: pC$
0%
$q_2 \: is \: positive, Q = 6 \: pC.$
0%
$q_2 \: is \: negative, Q = 6.25 \: pC$
0%
$q_2 \: is \: negative, Q = 6 \: pC.$

A nichrome wire of radius 0.321 mm and its length 2 m and 10 V potential difference across it. Find the current through it. (resistivity

$=15 \times 10^{-6}$ ohm.m )

0%
$2.2 A$
0%
$5.2 A$
0%
$1.08 A$
0%
$3.2 A$

Explanation

Ohm's law is the relationship between potential difference across a conductor and the current flowing through it.
Ohm's law states that "The current flowing through a conductor is directly proportional to the potential difference between its ends provided its temperature remains constant.
That is, $V\propto I\quad or\quad R=\frac { V }{ I }$
where R is a constant called resistance for a given metallic wire at a given temperature.
Experiment to verify ohm's law:
The circuit is set up as shown in the diagram below. Firstly, one cell is used and the current (I) is noted in the ammeter and the potential difference (V) in the voltmeter across the wire AB. Now, two cells, three cells and four cells are used and the observation is repeated. The readings in the ammeter and voltmeter are noted down for each repetition.
A graph is plotted between the current (I) and potential difference (V). The graph will be a straight line. This shows that the current flowing through a conductor is directly proportional to the potential difference across its ends.
That is,

$V\propto I\quad or\quad I\propto V\quad and\quad R=\frac { V }{ I }$ where R is the resistance of the metallic wire.
This proves for the ohm's law.

Conditions for which Ohm's law does not hold good:

Some materials have a temperature dependent specific resistance. So there are materials with a positive temperature coefficient (PTC) which increase resistance with increasing temperature and those with a negative temperature coefficient (NTC) which decrease resistance with increasing temperature. Hence, ohm's law does not hold for varying temperatures.
Ohm's law is all well and good for DC currents, but in AC circuits it needs a little expansion. This expansion is called impedance. It is a complex number in the form of $Z=R+j\times X$ , where the real part of the impedance is the ohmic resistance, and the imaginary part is called "reactance". If the reactance is positive, then you have coil-like, that is, inductive behaviour and if the reactance is negative you have capacitive behaviour.
Ohm's law cannot be applied to unilateral networks. A unilateral network has unilateral elements like diode, transistors, etc., which do not have same voltage current relation for both directions of current.
Ohms law is also not applicable for non linear elements. They are those which do not give current through it, is not exactly proportional to the voltage applied, that means the resistance value of those elements changes for different values of voltage and current. Examples of non linear elements are thyristor, electric arc, etc.

What is the shape of the equipotential surface for the line charge?

0%
Sphere
0%
cylinder
0%
will depend upon the type of charge.
0%
can't say.

Four equal charges

$+q$ are placed at four corners of a square with its centre at origin and lying in

$yz$ plane. The electrostatic potential energy of a fifth charge

$+q$ varies on x-axis as

Explanation

Now potential energy of charge +q at any distance $x$ from origin along x-axis is given

$U=\dfrac { 4k{ q }^{ 2 } }{ \sqrt { { x }^{ 2 }+{ { a }^{ 2 } }/{ 2 } } }$

$\dot { U } =-\dfrac { 8k{ qx }^{ 2 } }{ 2{ \sqrt { { x }^{ 2 }+{ { a }^{ 2 } }/{ 2 } } }^{ 3 } }$

Now as we can see for $x<0$ the slope potential energy curve is positive and at

$x=0$ the slope is zero and at

$x>0$ the slope is negative

Hence the correct answer is option (B)

Three point charges Q, 2Q and 8Q are placed on a straight line 9 cm long. Charges are placed in
such a way that the system has minimum potential energy. Then

0%
2Q and 8Q must be at the ends and Q at a distance of 3 cm from the 8Q.
0%
2Q and 8Q must be at the ends and Q at a distance of 6 cm from the 8Q.
0%
Electric field at the position of Q is zero.
0%
Electric field at the position of Q is $\frac{Q}{4\pi \varepsilon _{0}}$

Explanation

$\quad On\quad the\quad straight\quad line\quad the\quad charges\quad of\quad greater\quad magnitude\quad should\quad be\quad present\quad \\ at\quad the\quad opposite\quad ends\\ now\quad let\quad the\quad position\quad of\quad charge\quad Q\quad be\quad x\quad cm\quad distance\quad from\quad 8Q\quad charge\quad than\quad \\ distance\quad of\quad it\quad from\quad 2Q\quad becomes\quad 9-x\quad cm\\ The\quad expression\quad of\quad total\quad potential\quad energy\quad U\quad becomes:\\ \quad \quad \quad \quad \quad U=\quad \dfrac { K8{ Q }^{ 2 } }{ x } +\dfrac { K2{ Q }^{ 2 } }{ 9-x } ,\quad For\quad minimum\quad potential\quad energy\quad dU/dx=0\\ \qquad \qquad \dfrac { dU }{ dx } =\dfrac { -4 }{ { x }^{ 2 } } +\dfrac { 1 }{ { (9-x) }^{ 2 } } =0,\quad x=6,18\quad x=18\quad is\quad not\quad permissible\quad in\quad this\quad question\quad \\ \quad \quad Therefore\quad x=6\quad cm,\quad distance\quad of\quad Q\quad from\quad 8Q\quad is\quad 6cm\quad so\quad option\quad B\\ \quad Total\quad electric\quad field\quad E\quad at\quad x=6cm\quad due\quad to\quad 8Q\quad and\quad 2Q\quad point\quad charges\quad is\quad given\quad as:\\ E=\quad \dfrac { 8K{ Q }^{ 2 } }{ { 6 }^{ 2 } } -\dfrac { 2K{ Q }^{ 2 } }{ { 3 }^{ 2 } } =0\quad SO\quad option\quad C\quad$

The potential difference between points A and B, in a section of a circuit shown, is

0%
5 Volt
0%
1 Volt
0%
10 Volt
0%
-16Volt

Explanation

assuming the pot at A to be zero

$V_A-6-3-4+2-5=V_B$

$\Rightarrow V_B-V_A=-16$

which of the following is/are not proportional to the inverse square of the distance x?

0%
The potential at a distance x from an isolated point charge
0%
the electric field at a distance x from an isolated point charge
0%
The force per unit length between two thin, straight, infinitely long current carrying conductors, parallel to each other, separated by a distance x
0%
The gravitational attraction between two small bodies kept at a distance x apart

Explanation

We have to choose only those options which are

$\mathbf{not}$ proportional to the inverse square of the distance.

A-

$\dfrac{KQ}{r}$ Here, Potential is inversely proportional to distance not inverse square of the distance. So, this is correct option.

B-

$\dfrac{KQ}{r^{2}}$ Here, Electric Field is proportional to the inverse square of the distance. So, this is incorrect option.

C-

$\dfrac{\mu I_{1} I_{2}}{2\pi d}$ Here also, Force per unit length is inversely proportional to distance not inverse square of the distance. So, this is also a correct option.

D-

$\dfrac{GM_{1}M_{2}}{r^{2}}$ Here, Gravitational attraction is proportional to the inverse square of the distance. So, this is incorrect option.

Therefore, only (A) and (C) are correct options.

Positive charge flow from a body at ________to a body at _______. Fill in the blanks.

0%
higher potential, lower potential
0%
lower potential, higher potential
0%
higher charge, lower charge
0%
higher force, lower force

Electrical potential at the centre of a charged conductor is:

0%
zero
0%
twice as that on the surface
0%
half of that on the surface
0%
same as that on the surface

Explanation

Inside a charged conductor there is no electric field that is

$E=0$ .

And, we know that

$\dfrac{dV}{dr}=-E=0$

$\implies V=constant.$

Hence, electric potential at center of charged conductor is same as that on surface and at all points in conductor.

Answer_(D)

When two capacitors of capacities

$3\mu F$ and

$6\mu F$ are connected in series and connected to

$120 V,$ the potential difference across

$3\mu F$ is:

0%
$40V$
0%
$60V$
0%
$80V$
0%
$180V$

Explanation

$C=\dfrac {3\times 6}{9}=2\mu F$

$V=120 V$

$Q=CV=120\times 2=240\mu C$

$V_3=\dfrac {Q}{3\mu F}=\dfrac {240\mu C}{2\mu F}=80 V$

The two capacitors

$2\mu F$ and

$6\mu F$ are put in series, the effective capacity of the system in

$\mu F$ is:

0%
$8$
0%
$2$
0%
$\dfrac{3}{2}$
0%
$\dfrac{2}{3}$

Explanation

$2\mu F, 6\mu F$ are in series

$C_{eq}=\dfrac {C_1C_2}{C_1+C_2}$

$=\dfrac {2\times 6}{(2+6)}=\dfrac {12}{8}=\dfrac {3}{2}\mu F$

$4\times 10^{20}eV$ of energy is required to move a charge of 0.25 coulomb between two points, the p.d between them is:

0%
256 V
0%
512 V
0%
123 V
0%
215 V

Explanation

$\displaystyle V=\frac {W}{Q}=\frac {4\times 10^{20}\times 1.6\times 10^{-19}J}{0.25C}$

$=256 V$

$S_1$ and

$S_2$ are two equipotential surfaces on which the potentials are not equal. Then-

0%
$S_1$ and $S_2$ both cannot intersect.
0%
$S_1$ and $S_2$ both cannot be plane surfaces.
0%
In the region between $S_1$ and $S_2$ , the field is maximum where they are closed to each other.
0%
A line of force from $S_1$ and $S_2$ must be perpendicular to both.

Explanation

As we know that two equipotential surfaces can not intersect each other and lines of force are always perpendicular to the equipotential surface.
When two equipotential lines close together, the slope will be steep. So, close equipotential surfaces indicate a maximum electric field.

$(E=-\dfrac{dV}{dx})$
Ans:

$(A),(C),(D)$

The equation of an equipotential line in an electric field is

$y=2x$ , then the electric field strenght vector at

$(1,2)$ may be:

0%
$4\hat i+3\hat j$
0%
$4\hat i+8\hat j$
0%
$8\hat i+4\hat j$
0%
$-8\hat i+4\hat j$

Explanation

In this case, equipotential surface is a straight line having slope

$2$ ,
We know that at any point on equipotential surface, electric field should be perpendicular to the equipotential surface.

$(1,2)$ lies on our equipotential surface therefore field at this point should be perpendicular to the equipotential surface.
Hence our electric field will in the direction of line having slope

$(-1/2)$
our our field should make an angle

$\arctan { (-1/2) }$ with x- axis
which is true only in option (D) therefore option (D) is correct

From a supply of identical capacitors rated

$8\;\mu F, 250 \;V$ the minimum number of capacitors required to form a composite of

$16\;\mu F, 1000 \;V$ is

0%
2
0%
4
0%
16
0%
32

Explanation

The number of capacitance to be connected in series

$\displaystyle n=\frac{voltage \ rating \ required}{voltage\ rating \ of \ a \ capacitor \ given}=\frac{1000}{250}=4$
Equivalent capacitance,

$\displaystyle C_{eq}=(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})^{-1}=(\frac{4}{8})^{-1}=2$
Number of rows required

$\displaystyle =\frac{capacitance \ required}{capacitance \ of \ each \ row}=\frac{16}{2}$
Thus the minimum number of capacitors to be required

$=4\times 8=32$

Which of the following quantities are independent of the choice of zero potential or zero potential energy?

0%
Potential at a point.
0%
Potential difference between two points.
0%
Potential energy of a two charge system.
0%
Change in potential energy of a two charge system.

Explanation

$\delta V=V_A-V_B$

$=(V_A+V_{ref})-(V_B+V_{ref})=V_A-V_B$
i.e., does not depend on reference. Similarly,

$\delta U$ also independent of

$U_{ref}$ .

In a cathode ray oscillograph, the focussing of beam on the screen is achieved by

0%
Convex lenses
0%
Magnetic field
0%
Electric potential
0%
All of these

A condenser of capacity

$0.2\;\mu F$ is charged to a potential of 600 V. The battery is now disconnected and the condenser of capacity

$1\; \mu F$ is connected across it. The potential of the condenser will reduce to

0%
$600\ V$
0%
$300\ V$
0%
$100\ V$
0%
$120\ V$

Explanation

For (1), charge on condenser ,

$Q =C_1V_1=0.2\times 10^{-6} \times 600=12 \times 10^{-5} C$

For (2), now the equivalent capacitance of the circuit,

$C=C_1+C_2=0.2+1=1.2 \mu F$
potential drop of the condenser is

$V_2=\dfrac{Q}{C}=\dfrac{12 \times 10^{-5}}{1.2 \times 10^{-6}}=100 V$

Figure shows equi-potential surfaces for a two charges system. At which of the labeled points point will an electron have the highest potential energy?

0%
Point $A$
0%
Point $B$
0%
Point $C$
0%
Point $D$

Explanation

Potential energy when electron at B is $\displaystyle W_B=k\dfrac{(-e)(-q)}{r_b}=k\dfrac{qe}{r_b}$ where $r_b$ be the distance between $-e$ and $-q$ .

similarly, $\displaystyle W_C=k\dfrac{-qe}{r_c}$ and $\displaystyle W_D=k\dfrac{-qe}{r_d}$ and $W_A=0$

Thus, the highest potential energy point will be B.

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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