Solve

Guides

0

You visited us 0 times! Enjoying our articles? Unlock Full Access!

Question

The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is?
Proportional to the square root of the distance between the plates
Independent of the distance between the plates
Inversely proportional to the distance between the plates
Linearly proportional to the distance between the plates

A

Independent of the distance between the plates

B

Inversely proportional to the distance between the plates

C

Linearly proportional to the distance between the plates

D

Proportional to the square root of the distance between the plates

Open in App

Solution

Verified by Toppr

Since Electric field due to one plate of capacitor is given as

$E = \frac{σ}{2 ϵ_{o}}$

$E = \frac{Q}{2 A ϵ_{o}}$

Now force on another plate due to first plate will be given by

$F = Q E$

$F = \frac{Q^{2}}{2 ϵ_{o}}$

Hence, it is independent of distance between plates.

Was this answer helpful?

11

Similar Questions

Q1

The electrostatic force between the metal plates of an isolated parallel plate capacitor having a charge

Q

and area

A

is

View Solution

Q2

The electrostatic force between the metal plates of an isolated parallel plate capacitor having a charge

Q

and area

A

is

View Solution

Q3

Capacitance of a parallel plate capacitor is proportional to area of its plates and inversely proportional to distance between the plates.

View Solution

Q4

A parallel plate capacitor

C

has a charge

q

and potential

V

between the plates. Work required to double the distance between the plate is

View Solution

Q5

The distance between the plates of a reactangular parallel plate capacitor (PPC) of plate area

3140 {mm}^{2}

, whose capacitance is equal to that of an isolated metallic sphere of radius

2 m

will be

View Solution