Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 70

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Question 1
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The equivalent capacitance between x and y is:

A
$$\dfrac{5}{6}\mu F$$

B
$$\dfrac{7}{6}\mu F$$

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

D
$$4\mu F$$

Solution

We have,

Series capacitance, $$ \dfrac {1}{C_S} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + . . . + \dfrac{1}{C_n} $$

Parallel capacitance, $$ C_P = C_1 + C_2 + . . . + C_n $$
Question 2
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In the circuit shown in the figure, $$C1 = C, C_2 = 2C, C_3 = 3C, C_4 = 4C$$. Then, select the incorrect statement(s) from the following.

A
Maximum potential difference is across $$C_1$$

B
Minimum potential difference is across combination of $$C_3$$ and $$C_4$$

C
Maximum potential energy is in $$C_1$$

D
Minimum potential energy is in $$C_2$$

Solution

Using the laws for series combination and parallel combination of capacitors,

$$ \dfrac{1}{C_{net}} = \dfrac{1}{C} + \dfrac{1}{2C} + \dfrac{1}{(3C+4C)}$$

$$ \dfrac{1}{C_{net}} = 23/14C \Rightarrow C_{net} = 14C/23 $$

Charge provided by battery $$=14CV/23 $$

Therefore,

potential across $$C_1(C) = 14V/23 $$

potential across $$C_2(2C) = 7V/23 $$

potential across $$C_3 $$ and $$C_4 =2V/23 $$

also,

PE of $$C_1(C) = 2CV^2 \times (7/23)^2 $$

PE of $$C_2(2C) = CV^2 \times (7/23)^2 $$

PE of $$C_3(3C) = 6CV^2/23^2 $$

PE of $$C_4(4C) = 8CV^2/23^2 $$
Question 3
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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

A
$$\displaystyle\frac{Q'_1}{Q_1} = \displaystyle\frac{k+1}{k}$$

B
$$\displaystyle\frac{Q'_2}{Q_2} = \displaystyle\frac{k+1}{2}$$

C
$$\displaystyle\frac{Q'_2}{Q_2} = \displaystyle\frac{k+1}{2k}$$

D
$$\displaystyle\frac{Q'_1}{Q_1} = \displaystyle\frac{k}{2}$$

Solution

The capacitor 1 has capacitance $$C$$ and as dielectric contains in capacitor 2 so its capacitance becomes $$kC$$.
The net capacitance $$C_{eq}=\dfrac{C.kC}{C+kC}=\dfrac{Ck}{1+k}$$
and $$Q_{eq}=C_{eq}E$$
When dielectric is reemoved from capacitor 2, its capacitance becomes $$C$$.
now net capacitance $$C'_{eq}=\dfrac{C.C}{C+C}=\dfrac{C}{2}$$
and $$Q'_{eq}=C'_{eq}E=\dfrac{CE}{2}$$
When the capacitors are connected in series, they each have the same charge as the net capacitance. Thus, $$Q_{eq}=Q_1=Q_2$$ and $$Q'_{eq}=Q'_1=Q'_2$$
$$\therefore \dfrac{Q'_1}{Q_1}=\dfrac{Q'_2}{Q_2}=\dfrac{2k}{k+1}$$
Question 4
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Find the equivalent capacitance across AB (all capacitances in $$\mu$$ F) :

A
$$\dfrac {20}{3}\mu F$$

B
$$9 \mu F$$

C
$$48 \mu F$$

D
None of these

Solution

By law for series and parallel combinations of capacitors.
By wheatstone bridge
$${ C }_{ AB }=\left( \cfrac { 30\times 10 }{ 30+10 } +\cfrac { 6\times 2 }{ 6+2 } \right) \mu F$$
$${ C }_{ AB }=9\mu F$$
Question 5
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Figure (i) shows a capacitors of capacitance $$C_1$$. A dielectric of dielectric constant K filling half the volume between the plates are inserted as shown in (ii) and (iii) having capacitances $$C_2$$ and $$C_3$$. Then,

A
$$C_1=C_2=C_3$$

B
$$C_2<C_1\ and\ C_3>C_1$$

C
$$C_2>C_1 \space and \space C_3>C_1$$

D
$$C_2>C_1 \space and \space C_3< C_1$$

Solution

For (i), $$C_1=\dfrac{A\epsilon_0}{d}$$
For (ii), here two capacitors are in series- one is air filled capacitor with separation $$d/2$$ and other is dielectric filled with separation $$d/2$$ . Thus capacitance one is $$2C_1$$ and other is $$2kC_1$$ where k is dielectric constant.
$$C_2=\dfrac{2C_1.2kC_1}{2C_1+2kC_1}=\dfrac{2kC_1}{k+1}$$
For (iii), here two capacitors are in parallel- one is one is air filled capacitor with area $$A/2$$ and other is dielectric filled with area $$A/2$$.Thus capacitance one is $$C_1/2$$ and other is $$kC_1/2$$
$$C_3=C_1/2+kC_1/2=\dfrac{k+1}{2}C_1$$
As $$k>1, C_2>C_1$$ and $$C_3>C_1$$
Question 6
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Each side of a tetrahedral has a capacitor of capacitance $$C$$. Find the capacitance between point A and B.

A
$$\cfrac C2$$

B
$$2 C$$

C
$$C$$

D
$$\cfrac C3$$

Solution

The equivalent circuits are shown in figure.
Here , $$C_1=\dfrac{CC}{C+C}=C/2$$ and $$C_2=\dfrac{CC}{C+C}=C/2$$

Thus, $$C_{AB} =C/2+C/2+C=2C$$
Question 7
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A parallel-plate air capacitor of capacitance $$C_0$$ is connected to a cell of emf $$V$$ and then disconnected from it. A dielectric constant $$K$$, which can just fill the air gap of capacitor, is now inserted in it. Which of the following is incorrect?

A
The potential difference between the plates decreases $$K$$ times.

B
The energy stored in the capacitor decreases $$K$$ times.

C
The change in energy is $$\dfrac{1}{2}C_0\varepsilon ^2 (K-1)$$.

D
The change in energy is $$\dfrac{1}{2}C_0\varepsilon ^2 \left ( 1-\dfrac{1}{k} \right )$$.

Solution

As the capacitor is disconnected from cell so charge will remain constant .
here $$Q=C_0\epsilon$$
Before inserting dielectric potential difference between the plates is $$V_0=\dfrac{Qd}{A\epsilon_0}$$
after inserting, potential $$V=\dfrac{Qd}{Ak\epsilon_0}=\dfrac{V_0}{k}$$
before inserting dielectric the energy is $$U_0=\dfrac{Q^2}{2C_0}$$
after inserting, the energy is $$U=\dfrac{Q^2}{2kC_0}=\dfrac{U_0}{k}$$

Change in energy $$=U_0-U=\dfrac{Q^2}{2C_0}-\dfrac{Q^2}{2kC_0}=\dfrac{Q^2}{2C_0}(1-\dfrac{1}{k})=\dfrac{(C_0\epsilon)^2}{2C_0}(1-\dfrac{1}{k})=\dfrac{1}{2}C_0\epsilon^2(1-\dfrac{1}{k})$$
Question 8
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All the capacitance in the figure are of capacitance $$C$$. The effective capacitance between $$P$$ and $$Q$$ is :

A
$$2 C$$

B
$$3 C$$

C
$$4 C$$

D
$$1.5 C$$

Solution

Here first 3 capacitors are in parallel and then the 3 capacitors are also in parallel and thereafter the equivalent capacitances of both are in series as shown in figure below.
$$ C^1_{eq}=C+C+C=3C $$ and also $$C^2_{eq}=C+C+C=3C$$
$$C_{eq}=\dfrac{C^1_{eq}C^2_{eq}}{C^1_{eq}+C^2_{eq}}=\dfrac{9C^2}{6C}=\dfrac{3}{2}C=1.5C$$
Question 9
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Two identical capacitors are connected as shown in given figure, having a charge $$q_0$$. A dielectric slab is introduced between the plates of the capacitor (I) so as to fill the gap, keeping the battery remain connected. The charge on each capacitors now will be :

A
$$\displaystyle\frac{2q_0}{\left [ 1+(1/k) \right ]}$$

B
$$\displaystyle\frac{q_0}{\left [ 1+(1/k) \right ]}$$

C
$$\displaystyle\frac{2q_0}{(1+k)}$$

D
$$\displaystyle\frac{2q_0}{(1-k)}$$

Solution

Without dielectric : $$C_{eq}=\dfrac{C_0C_0}{C_0+C_0}=\dfrac{C_0}{2}$$ and $$Q_{eq}=\dfrac{C_0}{2}\times 2V_0=C_0V_0$$
As they are in series so the charge on each capacitors is equal to equivalent charge.i.e
$$Q_{eq}=q_0=C_0V_0$$
With dielectric in (I), the capacitance of it becomes $$kC_0$$
now $$\displaystyle C'_{eq}=\dfrac{kC_0C_0}{kC_0+C_0}=\dfrac{k}{k+1}C_0$$ and
$$\displaystyle Q'_{eq}=\dfrac{k}{k+1}C_0(2V_0)=\dfrac{2k}{k+1}q_0$$
Thus the on each capacitor is $$\displaystyle q'=Q'_{eq}=\dfrac{2k}{k+1}q_0=\dfrac{2q_0}{1+1/k}$$
Question 10
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For the configuration of media of permitivities $${ \varepsilon}_{o },\ \varepsilon $$ and $$\varepsilon_o$$ between parallel plates each of area $$A$$, as show in figure, the equivalent capacitance is :

A
$${ \varepsilon }_{ 0 } A/d$$

B
$${ \varepsilon }_{ 0 }\varepsilon A/d$$

C
$$\cfrac { { \varepsilon }_{ 0 }\varepsilon A }{ d(\varepsilon +{ \varepsilon }_{ 0 }) } $$

D
$$\cfrac { { \varepsilon }_{ 0 }\varepsilon A }{ (2\varepsilon +{ \varepsilon }_{ 0 }) } $$

Solution

Let the capacitance of the three capacitors be $${ C }_{ 1 },{ C }_{ 2 } \ \& \ { C }_{ 3 }$$ respectively,
$${ C }_{ 1 }=\cfrac { { \epsilon }_{ 0 }A }{ d } ,{ C }_{ 2 }=\cfrac { { \epsilon }A }{ d } \ \& \ { C }_{ 3 }=\cfrac { { \epsilon }_{ 0 }A }{ d } $$
Now, they are connected in series, So the equivalent capacitance is
$$\cfrac { 1 }{ C } =\cfrac { 1 }{ { C }_{ 1 } } +\cfrac { 1 }{ { C }_{ 2 } } +\cfrac { 1 }{ { C }_{ 3 } } $$
$$\Rightarrow \cfrac { 1 }{ C } =\cfrac { d }{ { \epsilon }_{ 0 }A } +\cfrac { d }{ \epsilon A } +\cfrac { d }{ { \epsilon }_{ 0 }A } $$
$$\cfrac { 1 }{ C } =\cfrac { d }{ A } \left[ \cfrac { 1 }{ \epsilon } +\cfrac { 2 }{ { \epsilon }_{ 0 } } \right] $$
$$\cfrac { 1 }{ C } =\cfrac { d\left( 2\epsilon +{ \epsilon }_{ 0 } \right) }{ A\epsilon { \epsilon }_{ 0 } } $$
$$\therefore C=\cfrac { { \epsilon }_{ 0 }\epsilon A }{ d\left( 2\epsilon +{ \epsilon }_{ 0 } \right) } $$