Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 32

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Weekly Quiz Competition

Question 1
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Seven capacitors each of capacitance $$2 \;\mu $$F are to be connected in a configuration to obtain an effective capacitance of $$\frac{10}{11}\mu F$$. which of the combination shown in figure will achieve the desired result

A

B

C

D

Solution

The capacitance in the individual cases are:
1. $$1/C_{net} = 1/10 + 1/2 + 1/2 \Rightarrow C_{net} = 10/11 $$
2. $$1/C_{net} = 1/8 + 1/2 + 1/2 +1/2 \Rightarrow 1/C_{net} = 1/8 + 3/2 = 13/8 $$
$$\therefore C_{net} = 8/13 $$.
3. $$1/C_{net} = 1/6 + 1/2 + 1/2 + 1/2 + 1/2 \Rightarrow 1/C_{net} = 1/6 + 2 = 13/6 $$
$$\therefore C_{net} = 6/13 $$
4. $$1/C_{net} = 1/4 + 5/2 = 11/4 \Rightarrow C_{net} = 4/11 $$
Question 2
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Condensers of capacities 2$$\mu $$F and 3$$\mu $$F are connected in series and a condenser of capacity 1$$\mu $$F is connected in parallel with them. The resultant capacity is :

A
$$0.83\mu $$F

B
$$6\mu $$F

C
$$2.2\mu $$F

D
$$0.54\mu $$F

Solution

The effective capacitance is,
$$ \dfrac{(2\times 3)}{(2+3)} \mu F + 1 \mu F = 2.2\mu F $$
Question 3
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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 :

A
$$2.4\times 10^{-4}coulomb$$

B
$$4.8\times 10^{-4}coulomb$$

C
$$7.2\times 10^{-4}coulomb$$

D
$$9.6\times 10^{-4}coulomb$$

Solution

We know when capacitor are connected in series charge present on them must be equal.
In seris $$\dfrac{1}{C_{s}}= \dfrac{1}{C_1}+\dfrac{1}{C_2}$$
$$\dfrac{1}{C_{s}}= \dfrac{1}{2}+\dfrac{1}{8}$$
$$\dfrac{1}{C_{s}}= \dfrac{5.}{8}$$
$$C_{s}= \dfrac{8}{5} \mu F$$
$$Q=CV$$
$$Q= \dfrac{8}{5}\times 10^{-6}\times 300$$
$$Q= 4.8\times 10^{-4}C.$$
Question 4
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5 condensers are connected as shown in the figure. The effective capacitance between ‘A’ and ‘B’ is :

A
$$\frac{25}{32}\mu F$$

B
$$\frac{32}{25}\mu F$$

C
$$\frac{32}{9}\mu F$$

D
$$\frac{9}{32}\mu F$$

Solution

we known
capacitors in series $$\dfrac{1}{C_s}=\dfrac{1}{c_1}+\dfrac{1}{c_2}+$$ - - -

capacitor in parallel $$C_p=c_1+c_2+$$ - - -

$$\dfrac{1}{C_s}=\dfrac{1}{2}+\dfrac{9}{32}$$
$$C_s=\dfrac{32}{25} \mu F$$
Question 5
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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 :

A
5.6times

B
11.2times

C
22.4 times

D
44.8 times

Solution

Initially,
$$ \ C=\dfrac{\epsilon_0\ A}{d}$$
$$\Rightarrow $$Distance is reduced by $$\frac{1}{4}$$ and K is added
$$\therefore C_{new}=\dfrac{k\ \epsilon_0\ A}{(^{d}/_{a})}$$
$$\Rightarrow C_{new}=2.8\times \dfrac{\epsilon_0\ A}{d}\times 4$$
$$\Rightarrow C_{new}=11.2\times C.$$
Question 6
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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 :

A
$$\frac{K}{2}C$$

B
2KC

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

D
(K+1)C

Solution

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=$$C_{eff}=\dfrac{KC}{2}+\dfrac{C}{2}=\dfrac{(K+1)C}{2}$$
Question 7
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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 :

A
6E, 6C

B
E, C

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

D
E, 6C

Solution

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 $$C_{1}=6C$$
Question 8
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The effective capacitance between ‘A’ and ‘B’ is :

A
$$\frac {1}{3}\mu F$$

B
$$44\mu F$$

C
$$7.6\mu F$$

D
$$6.7\mu F$$

Solution

Correct answer: Option A
Hint: Formula for effective capacitance in series is

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

Formula for effective capacitance in parallel is

$${C_{eq}} = {C_1} + {C_2}$$

Step 1: Finding Effective capacitance in series

the effective capacitance in the first section in series, is given by

$$\dfrac{1}{{{C_1}}} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}$$

$${C_1} = 1$$

Similarly,

$$\dfrac{1}{{{C_2}}} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}$$

$${C_2} = 1$$

$$\dfrac{1}{{{C_3}}} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}$$

$${C_3} = 1$$

$$\dfrac{1}{{{C_4}}} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}$$

$${C_4} = 1$$

Step 2: Finding Effective capacitance in parallel

Now $${C_1}$$ is in parallel with $$ {1} \mu F$$,Here $${C_5}$$ is net capacitance of $${C_5}$$ and $$1{\mu F}$$ capacitor.

$${C_5} = {C_1} + 1$$

$${C_5} = 1 + 1$$

$${C_5} = 2$$

Similarly,

$${C_6} = {C_2} + 1$$

$${C_6} = 1 + 1$$

$${C_6} = 2$$

$${C_7} = {C_3} + 1$$

$${C_7} = 1 + 1$$

$${C_7} = 2$$

$${C_8} = {C_4} + 1$$

$${C_8} = 1 + 1$$

$${C_8} = 2$$
And it is already given in figure that rest two capacitors are $${C_9}$$ and $${C_9}'$$ =.$$ {2} \mu F$$.

Step 3: Finding Resultant capacitance

Hence, the resultant capacitance in series given by,

$$\dfrac{1}{{{C_{AB}}}} = \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}$$

$${C_{AB}} = \dfrac{1}{3}\mu F$$

The effective capacitance is $$\dfrac{1}{3}\mu F$$
Option (A) is correct.
Question 9
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If a proton and an electron are accelerated through the same potential difference then:

A
both the proton and electron have same K.E

B
both the proton and electron have same momentum

C
both the proton and electron have same velocity

D
both the proton and electron have same temperature

Solution

P.D = V
K.E = eV
Velocity will be equal and opposite
Question 10
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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
$$\frac{q_{1}}{Q_{1}}=\frac{K+1}{K}$$

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

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

D
$$\frac{q_{2}}{Q_{2}}=\frac{K}{2}$$

Solution

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