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

Question

A wire of resistance 20 \(\Omega\) is divided into 10 equal parts. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _______ \(\Omega\).

Answer 1

To solve this question about the electrostatic force between the metal plates of a parallel plate capacitor, we need to explore the relationship between the force and the characteristics of the capacitor.

A parallel plate capacitor consists of two conductive plates separated by a distance, with equal and opposite charges on each plate. The electric field E between the plates of an ideal capacitor with charge Q and plate area A is given by:

E = \frac{Q}{\varepsilon_0 A}

where \varepsilon_0 is the permittivity of free space.

The force F on a charge q in an electric field E is:

F = q \cdot E

For a capacitor, the force F between the plates due to their charge is:

F = \frac{1}{2} \cdot \frac{Q^2}{\varepsilon_0 A}

Note that the above equation shows that the electrostatic force between the plates depends on the charge Q, permittivity \varepsilon_0, and area A but is independent of the distance between the plates.

Thus, the correct option among those given is independent of the distance between the plates.

Let's justify why the other options are incorrect:

Proportional to the square root of the distance between the plates: The formula does not include any term involving the square root of the distance.
Linearly proportional to the distance between the plates: The force equation lacks a term that is linearly related to the distance, as it is independent of such a distance.
Inversely proportional to the distance between the plates: There is no inverse distance term in the force equation.

Answer 2