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

Question

A 5-ohm resistor is connected to a 10 V battery. Calculate the current flowing through the resistor.

Answer 1

The question revolves around the electrostatic force between the plates of a parallel plate capacitor. To determine the correct answer, we need to analyze how this force is affected by the distance between the plates.

Let's consider a parallel plate capacitor with the following characteristics:

Capacitance: C
Charge: Q
Area of the plates: A
Distance between the plates: d

The electric field E between the plates of a capacitor is given by:

E = \frac{V}{d}

where V is the potential difference between the plates. At the same time, E is also given by:

E = \frac{\sigma}{\varepsilon_0}

where \sigma = \frac{Q}{A} is the charge density, and \varepsilon_0 is the permittivity of free space.

The electrostatic force F between the plates is:

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

It is important to note that the force F is independent of d, the distance between the plates, as long as the charge Q remains constant.

Hence, the correct answer is that the electrostatic force is independent of the distance between the plates.

Let's evaluate the given options based on our finding:

Inversely proportional to the distance between the plates: Incorrect. No term in the formula depends on d.
Independent of the distance between the plates: Correct. As derived, F does not depend on d.
Proportional to the square root of the distance between the plates: Incorrect. There is no square root relationship in the force calculation.
Linearly proportional to the distance between the plates: Incorrect. The force formula derived shows no dependence on the linear dimension d.

Answer 2