Question

A parallel plate capacitor has an electric field of \(10^5\) V/m between the plates. If the charge on the capacitor plate is 1 \(\mu\)C, the force on each capacitor plate is

• A. 0.5 N
• B. 0.05 N
• C. 0.005 N
• D. None of these

Hint:

Field due to one plate is E/2, so force is qE/2.

Answer 1

The Correct Option is B

To find the force on each capacitor plate, we can use the concept of the electric field between the plates of a parallel plate capacitor and the definition of the electric force.

The force \(F\) on a charged plate due to the electric field \(E\) is given by the formula:

\(F = qE\)

where:

• \(q\) is the charge on the plate,
• \(E\) is the electric field between the plates.

We are given:

• The electric field \(E = 10^5 \, \text{V/m}\),
• The charge on the plate \(q = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C}\).

Substituting these values into the formula:

\(F = (1 \times 10^{-6} \, \text{C})(10^5 \, \text{V/m})\)

\(F = 1 \times 10^{-6} \times 10^5\)

\(F = 0.1 \, \text{N}\)

However, the correct process should consider that only half of this field affects each plate due to symmetry. Half of 0.1 N is 0.05 N, explaining why the force is 0.05 N for each plate when considering the net force due to the field.

The correct answer is: \(0.05 \, \text{N}\).