Electrostatics Test

Result

Electrostatics Test - 64

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Weekly Quiz Competition

Question 1
4 / -1

Work done by an external agent is separating the parallel plate capacitor is

A
CV

B

C

D
None of these
Question 2
4 / -1

A parallel plate capacitor has an electric field of 10⁵ V/m between the plates. If the charge on the capacitor plate 1 µC, the force on each capacitor plate is

A
0.5 N

B
0.05 N

C
0.005 N

D
None of these

Solution
Question 3
4 / -1

A parallel capacitor has plate area A and separation d. It is charged to a potential difference V₀. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

A

B

C

D

Solution
Question 4
4 / -1

A thin metal plate P is inserted half way between the plates of a parallel plate capacitor of capacitance C in such a way that it is parallel to the two plates. The capacitance now becomes

A
C

B
C/2

C
4C

D
None of these
Question 5
4 / -1

If there are n capacitors in parallel connected to V volt source, then the energy stored is equal to

A
CV

B

C

D

Solution
Question 6
4 / -1

If n drops, each of capacitance C, coalesce to form a single big drop, then the ratio of the energy stored in the big drop to that in each small drop will be

A
n : 1

B
n^1/3 : 1

C
n^5/3 : 1

D
n² : 1

Solution
Question 7
4 / -1

64 small drops of mercury, each of radius r and charge q coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is

A
1 : 64

B
64 : 1

C
4 : 1

D
1 : 4

Solution
Question 8
4 / -1

Capacitance (in F) of a spherical conductor with radius 1 m is

A
1.1 × 10^–10

B
10^–6

C
9 × 10^–9

D
10^–3

Solution
Question 9
4 / -1

A parallel plate air capacitor is charged and then isolated. When a dielectric material is inserted between the plates of the capacitor, then which of the following does not change

A
Electric field between the plates

B
Potential difference across the plates

C
Charge on the plates

D
Energy stored in the capacitor
Question 10
4 / -1

Capacitance of a parallel plate capacitor becomes 4/3 times its original value if a dielectric slab of thickness t = d/2 is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is

A
8

B
4

C
6

D
2

Solution