Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 69

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Question 1
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Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be :(charge on each condenser is $$q_{0}$$; $$k$$ = dielectric constant )

A
$$\dfrac{2q_{0}}{1+1/k}$$

B
$$\dfrac{q_{0}}{1+1/k}$$

C
$$\dfrac{2q_{0}}{1+k}$$

D
$$\dfrac{q_{0}}{1+k}$$

Solution

When capacitors are connected in series the charge present on them must be equal.
$$\dfrac{Q}{c_{1}}+\dfrac{Q}{c_{2}}=2$$ ------------ (1)
Now when $$k$$ is introduced $$c_{2}$$ becomes $$kc_{2}$$

Let $$q_{o}$$ be the charge on capacitor
given $$c_{1}=c_{2}$$
$$\Rightarrow 2\dfrac { { q }_{ o } }{ c } =2,\dfrac { { q }_{ o } }{ c } =1$$ from (1)
$$ c ={ q }_{ o } $$

After introducing the dielectric, let charge on each plate become $$Q_{ 1 }$$
$$\Rightarrow \dfrac { Q_{ 1 } }{ { q }_{ o } } +\dfrac { Q_{ 1 } }{ k{ q }_{ 0 } } =2$$
$$\Rightarrow \dfrac { kQ_{ 1 }+Q_{ 1 } }{ k{ q }_{ o } } =2$$
$$\Rightarrow Q_{ 1 }=\dfrac { 2kq_{ o } }{ k+1 } $$
$$\Rightarrow Q_{1}=\dfrac{2q_{o}}{1+ \dfrac{1}{k}}$$
Question 2
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A parallel plate capacitor of capacity $$5 \mu F$$ and plate separation $$6 cm$$ is connected to a $$1V$$ battery and is charged. A dielectric of dielectric constant $$4$$ and thickness $$4 cm$$ is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is:

A
2$$\mu $$C

B
3$$\mu $$C

C
5$$\mu $$C

D
10$$\mu $$C

Solution
Question 3
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The capacity between the points A and B in the given circuit will be:

A
$$\dfrac{2C_{1}C_{2}+C_{3}(C_{1}+C_{2})}{C_{1}+C_{2}+2C_{3}}$$

B
$$\dfrac{C_{1}C_{2}+C_{2}C_{3}+C_{3}C_{1}}{C_{1}+C_{2}+C_{3}}$$

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

D
$$\dfrac{C_{1}C_{2}C_{3}}{C_{1}C_{2}+C_{2}C_{3}+C_{3}C_{1}}$$

Solution

The charge distribution is as shown in the figure.
Using Kirchhoff's law,
$$ V - q1/C_1 + (q2-q1)/C_3 - q1/C_1 =0 \Rightarrow V = 2q1/C_1 - (q2-q1)/C_3 $$
$$ V - q2/C_2 + (q1-q2)/C_3 - q2/C_2 =0 \Rightarrow V = 2q2/C_2 - (q1-q2)/C_3 $$
Therefore,
$$ C_1C_3 V = 2C_3q1 - C_2 q2 + C_2q1 = (2C_3+C_1)q1 - C_1q2$$
$$ C_2C_3 V = 2C_3q2 - C_2q1 + C_2q2 = (2C_3 + C_2)q2 - C_2q1$$
Solving these two equations we get,

$$q1 = \dfrac{C_1(C_2+C_3)V}{C_1+C_2+2C_3} $$

$$q2 = \dfrac{C_2(C_1 + C_3)V}{C_1+C_2+2C_3} $$

Equivalent capacitance $$=Q_{net}/V = (q1+q2)/V = \dfrac{2C_1C_2 + C_1C_3 + C_2C_3}{C_1 +C_2 + 2C_3}$$
Question 4
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The capacitance $$C_{AB}$$ in the given network is:

A
$$7\mu F$$

B
$$\dfrac{50}{7}\mu F$$

C
$$7.5\mu F$$

D
$$\dfrac{7}{50}\mu F$$

Solution

From the given circuit, we get $$\dfrac{C_1}{C_2} = \dfrac{5}{10} = \dfrac{1}{2}$$
And $$\dfrac{C_3}{C_4} = \dfrac{10}{5} = \dfrac{2}{1}$$
Thus $$\dfrac{C_1}{C_2}$$ is reciprocal of $$\dfrac{C_3}{C_4}$$
For this particular type of circuit, the equivalent capacitance is given by-
$$C_{net} = \dfrac{2C_1C_2 + C_3(C_1+C_2)}{C_1+C_2+2C_3} $$

$$C_{net} = \dfrac{2(5)(10) + 5(5+10)}{5+10+2(5)} = \dfrac{175}{25} = 7\mu F$$
Question 5
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The equivalent capacitance $$C_{AB}$$ of the circuit shown in the figure is :

A
$$\dfrac{5}{4}C$$

B
$$\dfrac{4}{5}C$$

C
$$2C$$

D
$$C$$

Solution

The given circuit can be reconstructed as shown (As the potential difference between A and B has to same through all possible paths)

Using the laws of combination of capacitors,

$$ C_{net} = C_1 + C_2 + C_3 \,where\,$$

$$ C_2 = \dfrac{C}{2} , \dfrac{1}{C_1} = \dfrac{1}{C} + \dfrac{1}{C} + \dfrac{1}{(C+ \dfrac{C}{2})} = \dfrac{2}{C} + \dfrac{2}{3C} = \dfrac{8}{3C} \Rightarrow C_1 = \dfrac{3C}{8} $$

Similarly, $$C_3 = \dfrac{3C}{8} $$

$$C_{net} = \dfrac{6C}{8} + \dfrac{C}{2} = \dfrac{5C}{4} $$
Question 6
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The effective capacity between A and B of the given network is

A
3C

B
2C

C
C

D
$$\frac{C}{3}$$

Solution
Question 7
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An air filled parallel plate capacitor has a capacitance $$1 pF$$. The separation between the plates is doubled and wax is inserted between the plates, then the capacitance becomes $$2 pF$$. What is the dielectric constant of wax ?

A
$$8$$

B
$$2$$

C
$$4$$

D
$$16$$

Solution

Capacitance of parallel plate capacitor with air ,$$C_a=\dfrac{\varepsilon _{0}A}{d} ....(1)$$

Capacitance of parallel plate capacitor with wax ,$$C_w=\dfrac{K\varepsilon _{0}A}{2d} .....(2)$$

$$(1)/(2),$$
$$ \dfrac{C_a}{C_w}=\dfrac{2}{K}$$

$$\dfrac{1}{2}=\dfrac{2}{K}$$

$$\therefore K=4$$
Question 8
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An isolated parallel plate capacitor is charged upto a certain potential difference. When a $$3mm$$ thick slab is introduced between the plates then in order to maintain the same potential difference, the distance between the plates is increased by $$2.4mm$$. Find the dielectric constant of the slab. (Assume charge remains constant)

A
5

B
10

C
2.5

D
7.5

Solution

$$C_f=\dfrac {\varepsilon_0A}{d'-t(1-\frac {1}{K})}$$
$$C_i=C_f$$
$$\dfrac {\varepsilon_0A}{d}=\dfrac {\varepsilon_0A}{d'-t(1-\dfrac {1}{K})}\Rightarrow d=d'-t(1-\dfrac {1}{K})$$
$$d'=d-2.4\times 10^{ -3 }$$
$$t=3mm$$
$$d=d+(2\cdot 4\times 10^{-3})-3\times 10^{-3}(1-\dfrac {1}{K})$$
$$K=5$$
Question 9
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The capacity of a parallel plate condenser without any dielectric is C. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant $$3$$, the capacity of the condenser becomes:

A
$$\dfrac {3}{4}C$$

B
$$\dfrac {9}{2}C$$

C
$$\dfrac {2}{3}c$$

D
$$\dfrac {3}{2}C$$

Solution

The capacitance of a parallel plate capacitor is given by

$$C = \dfrac {\varepsilon_o \varepsilon_r A}{d}$$,

where $$A$$ is the area of the plate and $$d$$ is the distance between them.

In the first case, there was no dielectric medium, i.e., $$\varepsilon_r = 1$$.

So we have $$C = \dfrac {\varepsilon_o A}{d}$$

In the second case, the distance is doubled, i.e., $$2d$$ and a substance of dielectric constant $$3$$ is inserted between the plates.

Hence capacitance will change to

$$C' = \dfrac {3 \varepsilon_o A}{2d}$$

So, $$ \dfrac{C'}{C} = \dfrac {3}{2}$$

$$ \implies C' = \dfrac {3}{2} C$$
Question 10
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A parallel plate capacitor is connected to a cell. It's positive plate A and the negative plate B have charges $$+Q$$ and $$-Q$$ respectively. A third plate C identical to A and B with charge $$+Q$$ is now introduced midway between the plates A and B parallel to them What is the charge on the inner surface of A ?

A
$$Q$$

B
$$\displaystyle \frac{\mathrm{Q}}{2}$$

C
$$\displaystyle \frac{3\mathrm{Q}}{2}$$

D
$$\displaystyle \frac{\mathrm{Q}}{4}$$

Solution

Let the capacitance of the A+B system be given by C = $$\dfrac{\epsilon A}{d}$$
Now when the plate is introduced in middle, the distance of A+C and C+B system is changed to $$\dfrac{d}{2}$$ So the capacitance will be $$\dfrac{2\epsilon A}{d}$$ = 2C
Now, Total charge in the system is Q assume A+C has q charge and C+B has Q+q charge
So, $$\dfrac{q}{2C}$$ + $$\dfrac{Q+q}{2C}$$ = $$\dfrac{Q}{C}$$
So, q = $$\dfrac{Q}{2}$$