Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 8

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Question 1
4 / -1

A parallel plate capacitor is connected across a 2V battery and charged. The battery is then disconnected and a glass slab is introduced between plates. Which of the following pairs of quantities decreases?

A
Charge and potential difference

B
Potential difference and energy stored

C
Energy stored and capacitance

D
Capacitance and charge

Solution

When battery is disconnected, charge remains constant. On introducing glass slab, capacity increases, Potential difference and energy stored decreases.
Question 2
4 / -1

A capacitor has some dielectric between its plates, and the capacitor is connected to a dc source. The battery is now disconnected and then the dielectric is removed, then

A
capacitance will increase.

B
energy stored will decrease.

C
electric field will increase.

D
voltage will decrease.

Solution

When the capacitor is connected to dc source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric capacitance decrease.
Energy stored (u = q²/2C) will increase.
Potential (V = q/C) will also increase.
Electric field (E = V/D) will increase.
Question 3
4 / -1

A capacitor of capacitance 700 pF is charged by 100 V battery. The electrostatic energy stored by the capacitor is

A
2.5 × 10⁻⁸J

B
3.5 × 10⁻⁶J

C
2.5 × 10⁻⁴J

D
3.5 × 10⁻⁴J

Solution

Here C = 700pF = 700 × 10⁻¹²F, V = 100V
Energy stored
Question 4
4 / -1

A 16 pF capacitor is connected to 70 V supply. The amount of electric energy stored in the capacitor is

A
4.5 × 10⁻¹²J

B
5.1 × 10⁻⁸J

C
2.5 × 10⁻¹²J

D
3.92 × 10⁻⁸J

Solution

Here C = 16pF = 16 × 10⁻¹²FV = 80V
Question 5
4 / -1

A capacitor is charged through a potential difference of 200 V, when 0.1 C charge is stored in it. The amount of energy released by it, when it is discharged is

A
5 J

B
10 J

C
20 J

D
2.5 J

Solution

Here, V = 200V, q = 0.1C
Energy stored,

When the capacitor is discharged, it releases the same amount of energy i.e., 10J
Question 6
4 / -1

A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is

A

B

C

D
ε₀E²Ad

Solution

Capacitance of a parallel plate capacitor is C = ε₀A/d ………(i)
Potential difference between the plates is
V = Ed
The energy stored in the capacitor is
Question 7
4 / -1

A metallic sphere of radius 18 cm has been given a charge of 5 × 10⁻⁶ C. The energy of the charged conductor is

A
0.2 J

B
0.6 J

C
1.2 J

D
2.4 J

Solution

Here r = 18cm = 18 × 10⁻²m, q = 5 × 10⁻⁶C

Energy of charged conductor is
Question 8
4 / -1

Two spherical conductors each of capacity C are charged to potential V and − V. These are then connected by means of a fine wire. The loss of energy will be

A
Zero

B

C
CV²

D
2CV²

Solution

Here, loss of energy
Question 9
4 / -1

Two condensers, one of capacity C and other of capacity C/2 are connected to a V-volt battery, as shown in the figure. The work done in charging fully both the condensers is

A
1/4 CV²

B
3/4 CV²

C
1/2 CV²

D
2CV²

Solution

As the capacitors are connected in parallel, therefore potential difference across both the condensors remains the same.

Work done in charging fully both the condensors is given by
Question 10
4 / -1

A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

A
zero

B

C

D
(K - 1)CV²

Solution

The potential energy of a charged capacitor before removing the dielectric slat is U = Q²/2C.
The potential energy of the capacitor when the dielectric slat is first removed and the reinserted in the gap between the plates is U = Q²/2C
There is no change in potential energy, therefore work done is zero.
Question 11
4 / -1

Two identical capacitors have the same capacitance C. One of them is charged to potential V₁ and the other to V₂. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

A

B

C

D

Solution

Initial energy of the combined system

On joining the two condensers in parallel, common potential.

∴ Final energy of the combined system

Decrease in energy
Question 12
4 / -1

Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio.

A
1 : 1

B
1 : 2

C
2 : 1

D
1 : 3

Solution

Half of the energy is dissipated during charging a capacitor.
Question 13
4 / -1

Two capacitors, 3μF and 4μF, are individually charged across a 6V6V battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the final total energy stored

A
1.26 × 10⁻⁴J

B
2.57 × 10⁻⁴J

C
1.25 × 10⁻⁶J

D
2.57 × 10⁻⁶J

Solution

As Q = CV
∴ Initially the charge on each capacitor is
Q₁ = C₁V₁ = (3μF)(6V) = 18μC
and Q₂ = C₂V₂= (4μF)(6V) = 24μC
When two capacitors is are joined to each other such that negative plate of one is attached with the positive plate of the other. The charges Q₁ and Q₂ are redistributed till they attain the common potential which is given by.

Common potential V = Total Charge/Total capacitance

Final energy stored,
Question 14
4 / -1

A parallel plate capacitor without any dielectric within its plates, has a capacitance C, and is connected to a battery of emf V. The battery is disconnected and the plates of the capacitor are pulled apart until the separation between the plates is doubled. What is the work done by the agent pulling the plates apart, in this process?

A

B

C

D
CV²

Solution

Capacitance of a parallel plate capacitor is C = ε₀A/d ……(i)
where A is the area of each plate and d is the distance between the plates. Initial energy stored in the capacitor,

When the separation between the plates is doubled , its capcaitance becomes

As the bettary is disconnected , so charged capacitor becomes isolated and charge on it will reamain constant ,
∴ Q = Q
C'V = C'V (AsQ = CV)

Final energy stored in the capacitor ,

Required work done,
Question 15
4 / -1

A series combination of n₁, capacitors, each of value C₁, is charged by a source of potential difference 4V. When another parallel combination of n₂, capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C₂, in terms of C₁ is then

A

B

C

D

Solution