The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

  1. independent of the distance between the plates.
  2. linearly proportional to the distance between the plates.
  3. inversely proportional to the distance between the plates.
  4. proportional to the square root of the distance between the plates.

Answer (Detailed Solution Below)

Option 1 : independent of the distance between the plates.
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Detailed Solution

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Explanation:

The electric force 'F' experienced by the charge 'q' is equal to the product of the magnitude of charge 'q' and the electric field 'E'.It is written as;

\(\vec F = Q \vec E\)

and electric field in isolated plate capacitor C is equal to;

\(\vec E = \frac {\sigma}{2\varepsilon _0}\)

Here, \(\sigma\) the surface charge density, \(\varepsilon _0\) is the permittivity of free space.

As we know that force experienced in the charge 'q' is,

\(\vec F = Q \vec E\)  ----(1)

where electric field,  \(\vec E = \frac {\sigma}{2\varepsilon _0}\)

Now, putting the value of an electric field in equation (1) we get;

\(\vec F = Q \frac {\sigma}{2\varepsilon _0}\) ----(2)

Now, surface charge density \(\sigma\) is written as;

\(\sigma = \frac {Q}{A}\)

Where Q is the charge and A is the area. Putting this value in equation (2) we get;

\(F= \frac{{{Q^2}}}{{2A{\varepsilon _0}}}\) ----(3)

In equation (3) we can see that the F is Independent of the distance between plates. 

Hence, option 1) is the correct answer.

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