Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The capacitance of a parallel plate capacitor with air as medium is $$6 \mu F$$. With the introduction of a dielectric medium, the capacitance becomes $$30\mu F$$. The permittivity of the medium is :
$$(\in_0= 8.85 \times 10^{-12} C^2N^{-1} m^{-2})$$

A
$$1.77 \times 10^{-12} C^2 N^{-1} m^{-2}$$

B
$$0.44 \times 10^{-10} C^2 N^{-1} m^{-2}$$

C
$$5.00 C^2 N^{-1} m^{-2}$$

D
$$0.44 \times 10^{-13} C^2 N^{-1} m^{-2}$$

Solution

Capacitance of air capacitor
$$C_0 = \dfrac{\varepsilon_0A}{d} = 6 \mu F$$ ......... (i)
When a dielectric of permittivity $$\varepsilon_r$$ and dielectric constant K is introduced between the plates, then
Capacitance, $$C = \dfrac{K \varepsilon_0 A}{d} = 30 \mu F$$ ..... (ii)
Dividing eq. (ii) by (i), we get
$$\dfrac{C}{C_0} = \dfrac{\dfrac{K\epsilon_o A}{ d}}{\dfrac{\varepsilon_0 A}{d}} = \dfrac{30}{6}$$
$$\Rightarrow K = 5$$
$$\therefore$$ permittivity of the medium
$$\varepsilon_0 = \varepsilon_0 K$$
$$= 8.85 \times 10^{-12} \times 5 = 0.44 \times 10^{-10}$$
Question 2
1 / -0

Two parallel metal plates having charges +Q and -Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will :

A
become zero

B
increase

C
decrease

D
remains same

Solution

Electric field between two parallel plates placed in vacuum is given by
E = $$\displaystyle \frac{\sigma}{\varepsilon_0}$$
In a medium of dielectric constant K, E' = $$\displaystyle \frac{\sigma}{\varepsilon_0 K}$$ For kerosene oil K > 1$$\Rightarrow$$ E' is less than E
Question 3
1 / -0

An electric charge $${10}^{-3}\mu C$$ is placed at the origin $$(0,0)$$ of $$(x,y)$$ co-ordinate system. Two points $$A$$ and $$B$$ are situated at $$(\sqrt {2}, \sqrt {2})$$ and $$(2,0)$$ respectively. The potential difference between the points $$A$$ and $$B$$ will be:

A
$$4.5\ V$$

B
$$9\ V$$

C
$$zero$$

D
$$2\ V$$

Solution

Given : $$q = 10^{-3} \mu C = 10^{-9} C$$

Distance of A from origin $$r_A = \sqrt{(\sqrt{2} - 0)^2 + (\sqrt{2} - 0)^2} = 2$$

Distance of B from origin $$r_B = 2$$

Thus potential due to charge at point A $$V_A = \dfrac{Kq}{r_A}$$ where $$K = \dfrac{1}{4\pi \epsilon_o} = 9 \times 10^9$$

$$\therefore$$ $$V_A = \dfrac{9 \times 10^9 \times 10^{-9}}{2} = 4.5$$ V

Potential due to the charge at B
$$V_B = \dfrac{Kq}{r_B} = \dfrac{9\times 10^9 \times 10^{-9}}{2} = 4.5$$ V

Potential difference between A and B
$$V_{AB} = V_A- V_B = 0$$
Question 4
1 / -0

A parallel plate capacitor with air as a dielectric has capacitance $$C$$. A slab of dielectric constant $$K$$, having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be:

A
$$(K + 3)\dfrac {C}{4}$$

B
$$(K + 2)\dfrac {C}{4}$$

C
$$(K + 1)\dfrac {C}{4}$$

D
$$\dfrac {KC}{4}$$

Solution

The two condensers with $$K$$ and with air are in parallel. With air, $$C_{1} = \dfrac {\epsilon_{0}}{d}\left (\dfrac {3A}{4}\right ) = \dfrac {3\epsilon_{0}A}{4d}$$
With medium, $$C_{2} = \dfrac {\epsilon_{0}K}{d}\left (\dfrac {A}{4}\right ) = \dfrac {\epsilon_{0}AK}{4d}$$
$$\therefore C' = C_{1} + C_{2}$$
or $$C' = \dfrac {3\epsilon_{0}A}{4d} + \dfrac {\epsilon_{0}AK}{4d} = \dfrac {\epsilon_{0}A}{d}\left [\dfrac {3}{4} +\dfrac {K}{4}\right ]$$
or $$C' = \dfrac {C}{4} (K + 3) \left [\because C = \dfrac {A\epsilon_{0}}{d}\right ]$$.
Question 5
1 / -0

A parallel plate capacitor of area $$'A'$$ plate separation $$'d'$$ is filled with two dielectrics as shown. What is the capacitance of the arrangement?

A
$$\dfrac {3K\epsilon_{0}A}{4d}$$

B
$$\dfrac {4K\epsilon_{0}A}{3d}$$

C
$$\dfrac {(K + 1)\epsilon_{0}A}{2d}$$

D
$$\dfrac {K(K + 3)\epsilon_{0}A}{2(K + 1)d}$$

Solution

$$c_{1} = \dfrac {(A/2)\epsilon_{0}}{d/2} = \dfrac {A\epsilon_{0}}{d}, c_{2} = K \dfrac {A\epsilon_{0}}{d}, c_{3} = K\dfrac {A\epsilon_{0}}{2d}$$
$$\therefore c_{eq.} = \dfrac {c_{1}\times c_{2}}{c_{1} + c_{2}} + c_{3} = \dfrac {(3 + K)KA\epsilon_{0}}{2d(K + 1)}$$
$$(\because C_{1}$$ and $$C_{2}$$ are in series and resultant of these two in parallel with $$C_{3})$$.
Question 6
1 / -0

Space between the plates of a parallel plate capacitor is filled with a dielectric slab. The capacitor is charged and then the supply is disconnected to it. If the slab is now taken out , then:

A
work is not done to take out the slab

B
energy stored in the capacitor reduces

C
potential difference across the capacitor is decreased

D
potential difference across the capacitor is increased

Solution

When a capacitor is charged and then the supply is disconnected charge present on it becomes constant (law of conservation of charge)
Initial capacitance $$C=\frac {k\varepsilon_0 A}{d}$$ (with slab)
Final capacitance $$C=\frac {\varepsilon_0 A}{d}$$ (slab removed)
We know $$Q= CV$$
As C decreases and Q remains constant V has to increase.
Question 7
1 / -0

A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,

A
potential increases

B
electric intensity incrases

C
energy decreases

D
capacity decreases

Solution

$$\rightarrow $$ When the capacitor is charged and battery is removed the charge present on capacitor remains constant. (Law of conservation of charge)
$$\rightarrow$$ We know $$C=\dfrac {\varepsilon_0 A}{d}$$
When slab is introduced, $$C=\dfrac {k\varepsilon_0 A}{d}$$
$$C$$ increases and $$Q$$ is constant.
We know: $$Q=CV$$
$$\therefore $$ $$V$$ has to decrease.
We know energy store in capacitor $$=\dfrac{1}{2} QV$$
$$\therefore $$ as V decreases, energy decreases.
Question 8
1 / -0

Fill in the blank:

In order to increase the capacity of a parallel plate condenser, one should introduce a sheet of $$\underline{\hspace{0.5in}}$$ between the plates (assume that the space is completely filled).

A
Mica

B
Tin

C
Copper

D
Stainless Steel

Solution

Dielectric constant is the factor by which the capacitance increases from its value in air when a dielectric is introduced between the plates.
Mica, among the materials mentioned above, is a dielectric material and hence to be used to increase the capacity.
Question 9
1 / -0

The capacity of a parallel plate air condenser is $$2\ \mu F$$. If a dielectric of dielectric constant $$4$$ is introduced between the plates, its new capacity is:

A
$$1.5\ \mu F$$

B
$$0.5\ \mu F$$

C
$$8\ \mu F$$

D
$$6\ \mu F$$

Solution

We know $$C=\dfrac{\epsilon_{0}A}{d}$$
When dielectric is added:
$$C'=\dfrac{K \epsilon_{0}A}{d}$$$$=4\times 2\ \mu F$$$$=8\mu F$$
Question 10
1 / -0

In an electrical circuit, which of the following quantities is analogous to temperature?

A
Potential

B
Resistance

C
Current

D
Charge

Solution

When there is a temperature difference, heat flows from high temperature to low temperature. Similarly, when there is a potential difference electric current flows from high potential to low potential. Hence, potential is analogous to temperature.

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Re-Attempt Weekly Quiz Competition

Electrostatic Potential and Capacitance Test - 23

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Electrostatic Potential and Capacitance Test - 23

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Question 1

$$1.77 \times 10^{-12} C^2 N^{-1} m^{-2}$$

$$0.44 \times 10^{-10} C^2 N^{-1} m^{-2}$$

$$5.00 C^2 N^{-1} m^{-2}$$

$$0.44 \times 10^{-13} C^2 N^{-1} m^{-2}$$

Solution

Question 2

become zero

increase

decrease

remains same

Solution

Question 3

$$4.5\ V$$

$$9\ V$$

$$zero$$

$$2\ V$$

Solution

Question 4

$$(K + 3)\dfrac {C}{4}$$

$$(K + 2)\dfrac {C}{4}$$

$$(K + 1)\dfrac {C}{4}$$

$$\dfrac {KC}{4}$$

Solution

Question 5

$$\dfrac {3K\epsilon_{0}A}{4d}$$

$$\dfrac {4K\epsilon_{0}A}{3d}$$

$$\dfrac {(K + 1)\epsilon_{0}A}{2d}$$

$$\dfrac {K(K + 3)\epsilon_{0}A}{2(K + 1)d}$$

Solution

Question 6

work is not done to take out the slab

energy stored in the capacitor reduces

potential difference across the capacitor is decreased

potential difference across the capacitor is increased

Solution

Question 7

potential increases

electric intensity incrases

energy decreases

capacity decreases

Solution

Question 8

Mica

Tin

Copper

Stainless Steel

Solution

Question 9

$$1.5\ \mu F$$

$$0.5\ \mu F$$

$$8\ \mu F$$

$$6\ \mu F$$

Solution

Question 10

Potential

Resistance

Current

Charge

Solution

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