A small object of mass $m$ has a charge $Q$ - AQA - A-Level Physics - Question 14 - 2019 - Paper 2

To find the potential difference $V$ between the plates, we need to consider the forces acting on the charged object.

1. Understand the Forces: The charged object will experience an electric force due to the electric field $E$ between the plates. The electric force can be expressed as: $F = Q imes E$

2. Relate Electric Field to Potential Difference: The electric field $E$ between two plates separated by a distance $d$ and having a potential difference $V$ is given by: $E = \frac{V}{d}$

3. Set Forces Equal: Since the object is stationary, the electric force must balance the gravitational force acting on it. The gravitational force is: $F_g = m imes g$ where $g$ is the acceleration due to gravity.

4. Equate Forces: Setting the electric force equal to the gravitational force gives: $Q \times \frac{V}{d} = m \times g$

5. Solve for $V$: Rearranging the above equation to solve for $V$ results in: $V = \frac{m \times g \times d}{Q}$

From the provided options, match this equation to the choices given. Thus, the correct response is:

Answer: The expression for $V$ is given by option C: $\frac{m \times g \times d}{Q}$.