Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The capacitance of a capacitor becomes $$\dfrac{7}{6}$$ times its original value if a dielectric slab of thickness, $$t=\dfrac{2}{3}d$$ is introduced in between the plates. d is the separation between the plates. The dielectric constant of the dielectric slab is :

A
$$\dfrac{14}{11}$$

B
$$\dfrac{11}{14}$$

C
$$\dfrac{7}{11}$$

D
$$\dfrac{11}{7}$$

Solution

Now $$C_{eff}$$ is series combination of $$\dfrac{l_{0}A}{x}, \dfrac{l_{0}A}{d-t-x}\ and\ \dfrac{kl_{0}A}{t}$$

$$\dfrac{1}{C_{eff}}=\dfrac{x}{l_{0}A}+\dfrac{d-t-x}{l_{0}A}+\dfrac{t}{kl_{0}A}$$

$$\Rightarrow C_{eff}=\dfrac{kl_{0}A}{(d-t)k+t}$$

$$\Rightarrow$$ given $$C_{eff}=\dfrac{7}{6}\cdot \dfrac{l_{0}A}{d}$$

$$\dfrac{7}{6}\cdot \dfrac{l_{0}A}{d}=\dfrac{kl_{0}A}{(d-\dfrac{2}{3}d)k+\dfrac{2}{3}d}$$

$$\Rightarrow \dfrac{7}{6}=\dfrac{3k}{(\dfrac{1}{3})k+\dfrac{2}{3}}$$

$$\Rightarrow 7k+14=18k$$

$$\Rightarrow k=\dfrac{14}{11}$$
Question 2
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In the capacitor of capacitance $$20\ \mu F$$, the distance between plates is $$2\ mm$$. If a material of dielectric constant $$ 2$$ is inserted between the plates, then the capacitance of the system is :

A
$$20 \mu F$$

B
$$30 \mu F$$

C
$$22 \times 5 \mu F$$

D
$$40 \mu F$$

Solution

We know
$$C=\dfrac{\epsilon_{0}A}{d}$$
When k is added
$$C_{1}=\dfrac{k\epsilon_{0}A}{d}$$
$$\Rightarrow C_{1}=2\times 20$$
$$\Rightarrow C_{1}=40 \mu F$$
Question 3
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A highly conducting sheet of aluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates at distance $$\dfrac{d}{2}$$ from positive plate where $$d$$ is distance between plates. If the capacitance before the insertion of foil was $$10 \; \mu F$$ , its value after the insertion of foil will be:

A
$$20\; \mu F$$

B
$$10 \;\mu F$$

C
$$5 \;\mu F$$

D
Zero

Solution

Initial capacitor $$C=\dfrac{\epsilon_{0}A}{d}=10\mu R$$
After inserting aluminium foil the effective capacitor is series combination of $$C_{1}=C_{2}=\frac{\epsilon_{0}A}{\frac{d}{2}}$$
$$\therefore \dfrac{1}{C_{eff}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$$
$$\Rightarrow C_{eff}=\dfrac{C_{1}}{2}$$
$$\Rightarrow C_{eff}=\dfrac{\epsilon_{0}A}{d}$$
$$\Rightarrow C_{eff}=10\mu F$$
Question 4
1 / -0

Two metal plates are separated by a distance $$d$$ in a parallel plate condenser. A metal plate of thickness $$t$$ and of the same area is inserted between the condenser plates. The value of capacitance increases by a factor of :

A
$$\dfrac{d-t}{d}$$

B
$$\left ( 2-\dfrac{t}{d} \right )$$

C
$$\left ( t-\dfrac{t}{d} \right )$$

D
$$\dfrac{1}{\left ( 1-\dfrac{t}{d} \right )}$$

Solution

Hint:
Capacitance of a parallel plate capacitor with distance $$d$$ between the plates is given by,
$$C=\dfrac{A{{E}_{o}}k}{d}$$
Where,
$$A$$ is area of plates,
$${{E}_{o}}$$ is permittivity of free space,
$$k$$ is dielectric constant of medium filled between plates.

Capacitance of a parallel plate capacitor with distance $$d$$ between the plates, with a dielectric slab of thickness $$t$$ and dielectric constant$$k$$ between the plates is given by,
$$C=\dfrac{A{{E}_{o}}}{d-t\left( 1-\dfrac{1}{k} \right)}$$
Note that for metal, value of $$k$$ is infinite.

Step 1: Given.
A parallel plate capacitor with distance d between plates of area A. Capacitance is given by,
$$C=\dfrac{A{{E}_{o}}}{d}$$

Step 2: Calculate new capacitance in terms of old capacitance.
A metal plate of thickness t is inserted between them. New capacitance is:
$$C'=\dfrac{A{{E}_{o}}}{d-t\left( 1-\dfrac{1}{k} \right)}$$
$$\Rightarrow C'=\dfrac{A{{E}_{o}}}{d-t}$$ (Since, k is infinite for metal)
$$\Rightarrow C'=\dfrac{A{{E}_{o}}}{d(1-\dfrac{t}{d})}$$
$$\Rightarrow C'=\dfrac{C}{(1-\dfrac{t}{d})}$$
$$\Rightarrow C'=\dfrac{1}{(1-\dfrac{t}{d})}C$$
Thus value of capacitance increases by a factor $$\dfrac{1}{(1-\dfrac{t}{d})}$$

Option D is correct.
Question 5
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An oil condenser has a capacity of $$100\; \mu F$$ . The oil has dielectric constant 2. When the oil leaks out , its new capacity is :

A
$$200\; \mu F$$

B
$$0.02\; \mu F$$

C
$$50 \;\mu F$$

D
$$0.5 \;\mu F$$

Solution

We know,
$$C=\dfrac{\epsilon_{0}A}{d}$$
When oil was present $$\epsilon_{0}$$ becomes $$k\epsilon_{0}$$
$$\therefore$$ when oil leaks out capacitors decreases by k.
$$\therefore C_{leaks}=\dfrac{c}{k}$$
$$=\dfrac{100\mu R}{2}$$
$$C_{leaks}=50\mu F$$
Question 6
1 / -0

A positive point charge q is carried from a point B to a point A in the electric field of a point charge +Q. If the permittivity of free space is $$\epsilon _{0}$$ , the work done in the process is given by

A
$$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a}-\frac{1}{b} \right ]$$

B
$$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a}+\frac{1}{b} \right ]$$

C
$$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a^{2}}-\frac{1}{b^{2}} \right ]$$

D
$$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a^{2}}+\frac{1}{b^{2}} \right ]$$

Solution

Work done, $$W=q[V_2-V_1]$$
Voltage at $$a, V_a=\dfrac {kQ}{a}$$
Voltage at $$b, V_b=\dfrac {kQ}{b}$$
$$\Rightarrow W=q[\dfrac {kQ}{a}-\dfrac {kQ}{b}]$$
$$=\dfrac {qQ}{4\pi \varepsilon_0} \left[ \dfrac {1}{a}-\dfrac {1}{b} \right]$$
Question 7
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A capacitor of $$10 \mu F$$ capacitance is charged by a $$12 V$$ battery. Now the space between the plates of capacitors is filled with a dielectric of dielectric constant $$K = 3$$ and again it is charged. The magnitude of the charge is :

A
$$120 \mu C$$

B
$$240 \mu C$$

C
$$360 \mu C$$

D
$$480 \mu C$$

Solution

$$\textbf{Step 1: Initial charge [Refer Fig. 1]} $$

$$Q_1 = C_1 V_1 = 10\ \mu F \times 12\ V $$

$$ = 120\ \mu C$$

$$\textbf{Step 2: Final capacitance after insertion of dielectric [Refer Fig. 2]} $$

$$C_{2} = \dfrac{k \epsilon_0 A}{d} = kC_{1} = 3 \times 10\ \mu F$$

$$C_{2} = 30\ \mu F $$

$$\textbf{Step 3: Final charge [Refer Fig. 3]}$$

As the battery remains connected, voltage across capacitor remains same.

i.e., $$V_{2} = V_{1} = 12\ V $$

$$Q_{2} = C_{2} V_{2} $$

$$Q_{2} = 30\ \mu F \times 12\ V $$

$$ = 360\ \mu C $$
Hence, option C is correct.
Question 8
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The capacity of a parallel plate condenser is $$10 \;\mu F$$ without the dielectric. Material with a dielectric constant of $$2$$ is used to fill half-thickness between the plates. The new capacitance is $$\underline{\hspace{0.5in}} \;\mu F$$ :

A
$$10$$

B
$$20$$

C
$$15$$

D
$$13.33$$

Solution

Initial capacitance $$=\dfrac{\epsilon_{0}A}{d}=10\mu F$$
Now effective capacitance is series combination of $$\dfrac{\epsilon_{0}A}{\dfrac{d}{2}}$$ and $$\dfrac{k\epsilon_{0}A}{\dfrac{d}{2}}$$
$$\therefore C_{eff}=$$ series of $$20\mu F$$ and $$40\mu F$$
$$\dfrac{1}{C_{eff}}=\dfrac{1}{20}+\dfrac{1}{40}$$
$$C_{eff}=13.33 \mu F$$
Question 9
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The plates of a parallel plate capacitor are charged up to $$200$$ volts. A dielectric slab of thickness 4mm is inserted between the plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $$3.2mm$$. The dielectric constant of a dielectric slab is :

A
$$1$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Initial voltage$$ =200 V$$
$$q=C.V$$
$$q=\dfrac{\epsilon_0.A}{d}V$$ (1)
When dielectric is inserted $$q$$ remains constant and V decreases by k.
$$q=\dfrac{\epsilon_0A}{d'+t-\dfrac{t}{k}}V$$ (2)
From equation 1 and 2 we have,
$$d=d'+t-\dfrac{t}{k}$$
$$d-d'=1.6mm=2-\dfrac{2}{k}$$
solving,
$$k=5$$
Question 10
1 / -0

A number of identical condensers are first connected in parallel and then in series. The equivalent capacitance are found to be in the ratio $$9:1$$. The number of condensers used is :

A
$$9$$

B
$$81$$

C
$$3$$

D
$$100$$

Solution

In series
$$\dfrac{1}{C_{eff}}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}+$$ upto n terms
$$C_eff=\dfrac{C}{n}$$

In parallel
$$C_eff=C+C+$$ upto n terms

$$C_eff=nC$$

$$\dfrac{C_{P}}{C_{S}}=\dfrac{9}{1}$$

$$\dfrac{nC}{\dfrac{C}{n}}=\dfrac{9}{1}$$

$$n^{2}=9$$

$$n=3$$

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Electrostatic Potential and Capacitance Test - 29

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Electrostatic Potential and Capacitance Test - 29

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Question 1

$$\dfrac{14}{11}$$

$$\dfrac{11}{14}$$

$$\dfrac{7}{11}$$

$$\dfrac{11}{7}$$

Solution

Question 2

$$20 \mu F$$

$$30 \mu F$$

$$22 \times 5 \mu F$$

$$40 \mu F$$

Solution

Question 3

$$20\; \mu F$$

$$10 \;\mu F$$

$$5 \;\mu F$$

Zero

Solution

Question 4

$$\dfrac{d-t}{d}$$

$$\left ( 2-\dfrac{t}{d} \right )$$

$$\left ( t-\dfrac{t}{d} \right )$$

$$\dfrac{1}{\left ( 1-\dfrac{t}{d} \right )}$$

Solution

Question 5

$$200\; \mu F$$

$$0.02\; \mu F$$

$$50 \;\mu F$$

$$0.5 \;\mu F$$

Solution

Question 6

$$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a}-\frac{1}{b} \right ]$$

$$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a}+\frac{1}{b} \right ]$$

$$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a^{2}}-\frac{1}{b^{2}} \right ]$$

$$\displaystyle \frac{qQ}{4\pi \epsilon _{0}}\left [ \frac{1}{a^{2}}+\frac{1}{b^{2}} \right ]$$

Solution

Question 7

$$120 \mu C$$

$$240 \mu C$$

$$360 \mu C$$

$$480 \mu C$$

Solution

Question 8

$$10$$

$$20$$

$$15$$

$$13.33$$

Solution

Question 9

$$1$$

$$4$$

$$5$$

$$6$$

Solution

Question 10

$$9$$

$$81$$

$$3$$

$$100$$

Solution

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