Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

If 3 capacitors of values 1, 2 and $$3 \mu F$$ are available. The maximum and minimum values of capacitance one can obtain by different combinations of the three capacitors together are respectively:

A
$$6\mu F, \frac{6}{11}\mu F$$

B
$$6\mu F\frac{11}{6}\mu F$$

C
$$3\mu F, 1 \mu F$$

D
$$4\mu F, 2 \mu F$$

Solution

minimum capacity is obtained when they are in series
$$\dfrac{1}{C_{eff}}=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{11}{6}$$
$$\therefore C_{eff}=\dfrac {6}{11}\mu F$$
maximum capacity is obtained when they are in parallel
$$C_{eff} =1+2+3$$
$$C_{eff}=6 \mu F$$
Question 2
1 / -0

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 :

A
$$22 \mu C$$

B
$$100 \mu C$$

C
$$2800 \mu C$$

D
$$200 \mu C$$

Solution

$$\dfrac{1}{C_{ef}}=\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{15}$$
$$\dfrac{1}{C_{ef}}=\dfrac{1}{3}+ \dfrac{5}{30}=\dfrac{3}{6}$$
$$\therefore C_{eff} = 2$$
$$\therefore$$ 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
$$\therefore charge \ on \ 15 \mu \ F \ is \ 200 \mu C$$
Question 3
1 / -0

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

A
$$10^{-5}J$$

B
$$2 \times 10^{-5}J$$

C
$$3 \times 10^{-5}J$$

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

Solution

Given energy = $$1\times 10^{-5}\ $$
When battery is connected, $$V$$ across capacitor will be constant.
When dielectric is filled, $$C$$ increases by $$K$$
$$\therefore New\ energy\ = \dfrac{1}{2}(KC) V\ ^{2}$$
$$=K\dfrac{1}{2} C V\ ^2 ( Given: \dfrac{1}{2} C V^2 = 1 \times 10^{-5} \ J)$$
$$= K (1 \times 10^{-5})\ J$$
$$= 5 \times (1 \times 10^{-5}) \ J$$
$$= 5 \times 10 ^{-5} \ J$$
Question 4
1 / -0

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 :

A
0.1 J

B
2.5 J

C
0.4 J

D
5 J

Solution

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

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

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

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

$$= 0.4J$$
Question 5
1 / -0

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

A
$$60\; V, 40 \; V$$

B
$$70\; V, 30\; V$$

C
$$75\; V, 25\; V$$

D
$$80 \; V, 20 \; V$$

Solution

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 $$=80V,20V$$
Question 6
1 / -0

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

A
$$22.25\; pF$$

B
$$20.02 \;pF$$

C
$$17.8\; \mu F$$

D
$$17.8 \;pF$$

Solution

when area of plate are different we consider the area which is smaller to calculate capacitor,
$$\therefore c=\dfrac{\epsilon_\circ{}\ A}{d}$$

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

$$ \Rightarrow c=17.8\ pF$$
Question 7
1 / -0

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:

A
$$\sqrt{2}:1$$

B
2 :1

C
1:1

D
$$1:\sqrt{2}$$

Solution

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

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

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

$$\dfrac{C_{2}}{C_{1}}=\dfrac{\dfrac{2 {\epsilon}_{0}A}{d}}{\dfrac{{\epsilon}_{0}A}{d}}=2:1$$
Question 8
1 / -0

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

A
$$\dfrac{1}{4}$$

B
$$4$$

C
$$3$$

D
$$1$$

Solution

$$\textbf{Step 1 - Writing capacitance equation}$$
$$C = \dfrac {\varepsilon_{0}A}{d}$$ $$....(1)$$
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 $$4$$.
Question 9
1 / -0

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)$$q_{0}=240\mu C$$

A
A, C are correct

B
A, B are correct

C
B, D are correct

D
C, D are correct

Solution

When capacitor are connected is series charge present on capacitor is same.
Now,
$$ q_0 =CV=240\times 10^{-6}$$
We know,
$$V_0=\dfrac{q_0}{C}$$
$$V_0=\dfrac{{240}\times 10^{-6}}{3\times 10^{-6}}$$
$$V_0=80V$$
Question 10
1 / -0

Given four capacitors each of capacity 12$$\mu $$F. To get a capacity of 9$$\mu $$F, what combination can be used:

A
All in series

B
All in parallel

C
3 in parallel and 1 series with them

D
2 in parallel and 2 in series

Solution

Formula for equivalent capacitance in parallel
$$C_{eq}= C_1+ C_2$$
Formula for equivalent capacitance in series
$$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$$
So, For 3 capacitance $$( 9 \mu Feach)$$ in parallel and 1 series ,the equivalent capacitance =
$$\dfrac{1}{C_{eq}} = \dfrac{1}{36} + \dfrac{1}{12}$$
$$ \Rightarrow C_{eq}= \dfrac{36 \times 12}{36+12}$$

$$ \Rightarrow C_{eq}= 9 \mu F$$
Therefore, C is correct option.

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Electrostatic Potential and Capacitance Test - 31

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Electrostatic Potential and Capacitance Test - 31

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Question 1

$$6\mu F, \frac{6}{11}\mu F$$

$$6\mu F\frac{11}{6}\mu F$$

$$3\mu F, 1 \mu F$$

$$4\mu F, 2 \mu F$$

Solution

Question 2

$$22 \mu C$$

$$100 \mu C$$

$$2800 \mu C$$

$$200 \mu C$$

Solution

Question 3

$$10^{-5}J$$

$$2 \times 10^{-5}J$$

$$3 \times 10^{-5}J$$

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

Solution

Question 4

0.1 J

2.5 J

0.4 J

5 J

Solution

Question 5

$$60\; V, 40 \; V$$

$$70\; V, 30\; V$$

$$75\; V, 25\; V$$

$$80 \; V, 20 \; V$$

Solution

Question 6

$$22.25\; pF$$

$$20.02 \;pF$$

$$17.8\; \mu F$$

$$17.8 \;pF$$

Solution

Question 7

$$\sqrt{2}:1$$

2 :1

1:1

$$1:\sqrt{2}$$

Solution

Question 8

$$\dfrac{1}{4}$$

$$4$$

$$3$$

$$1$$

Solution

Question 9

A, C are correct

A, B are correct

B, D are correct

C, D are correct

Solution

Question 10

All in series

All in parallel

3 in parallel and 1 series with them

2 in parallel and 2 in series

Solution

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