Question
Download Solution PDFThe electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
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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