Question

Distance between plates of a parallel plate charged capacitor (disconnected from battery) is increased. The quantity that remains constant is
a.Potential difference between plate
b. Charge
c. Electric field between plates
d. Both (2) and (3)​

Answer 1

Answer:

Correct option is

B

Potential difference across the capacitor

C

Energy density between the plates

Since the charge always remains conserved in al isolated system, it will remain the same

Now,

V=

ε

0

A

Qd

Here, Q, A, and d are the charge, area, and distance between the plates respectively.

Thus as d increases, V increases.

Energy is given by

E=

2

qV

So.it will also increase with increase in the value of the potential.

Energy density u, that is, energy stored per unit volume in the electric field is given by

u=

2

1

ε

0

E

2

So, u will remain constant with increase in distance between the plates