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**Q.1. In a certain region of space with volume 0.2 m ^{3}, the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is: (2020) **

given v = const. (5 volt)

E = 0

C

30μF = K6μF

K = 5

∴ ε = ε

ε = 44.25 × 10

ε = 0.4425 × 10

ε = 0.44 × 10

A: 2.5 A

B: 25.1 A

C: 1.7 A

D: 2.05 A

Ans:

A: independent of the distance between the plates.

B: linearly proportional to the distance between the plates

C: proportional to the square root of the distance between the plates.

D: inversely proportional to the distance between the plates.

Ans:

Solution:

F = QE

Charge on capacitor

q = CV

when it is connected with another uncharged capacitor.

A: Maximum work is required to move q in figure (c)

B: Minimum work is required to move q in figure (a)

C: Maximum work is required to move q in figure (b)

D: In all the four cases the work done is the same

Ans:

W = qΔV

as ΔV is same in all conditions, work will be same.

B: 0%

C: 20%

D: 75%

Ans:

Solution:

Consider the figure given above.

When switch S is connected to point 1, then initial energy stored in the capacitor is given as,

When the switch S is connected to point 2, energy dissipated on connection across 8μF will be,

Therefore, per centage loss of energy

A: The charge on the capacitor is not conserved.

B: The potential difference between the plates decreases K times

C: The energy stored in the capacitor decreases K times

Ans:

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it.

The charge on the capacitor is given by

Q = CV

The energy stored in the capacitor is

E = 1/2 CV

When a dielectric slab of dielectric constant K is inserted in it, the charge Q is conserved.

The capacitance becomes K times the original capacitance. (C' = KC)

The voltage becomes 1/K time the original voltage

V' = V/K

The change in energy stored is

A: 24N

B:

C:

D: 30N

Ans:

A:

B: Both are zero

C:

D:

Ans:

Electric field inside, E

Potential, V

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