Question

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be

• A. 5 N
• B. 10 N
• C. 20 N
• D. Zero

Answer 1

The Correct Option is A

To solve this problem, we need to understand how the electric force on a charged particle changes when one plate of a charged capacitor is removed.

Initially, when both plates of the capacitor are present, the electric field between the plates exerts a force on a charged particle placed between them. This electric field E is given by:

E = \frac{V}{d}

where V is the potential difference between the plates, and d is the distance between the plates.

The force F on a charged particle with charge q is given by:

F = qE

Given that the force initially is 10 N, we have:

F_1 = q \cdot \frac{V}{d} = 10 \, \text{N}

Now, if one plate of the capacitor is removed, the configuration becomes equivalent to a single plate, and the electric field due to a single charged plate is halved compared to the field between two plates. Hence, the new electric field E' becomes:

E' = \frac{E}{2}

The new force F_2 acting on the charged particle is:

F_2 = qE' = q \cdot \frac{V}{2d}

Substituting the initial conditions:

F_2 = \frac{1}{2} \cdot q \cdot \frac{V}{d} = \frac{1}{2} \times 10 \, \text{N} = 5 \, \text{N}

Therefore, the force acting on the particle after removing one of the capacitor plates will be reduced to half, becoming 5 N.

Thus, the correct answer is 5 N.