Question

In parallel plate capacitor, electric field between the plates is \( E \). If the charge on the plates is \( Q \), then the force on each plate is

• A. \( QE \)
• B. \( \frac{QE}{2} \)
• C. \( QE^2 \)
• D. \( \frac{QE^2}{2} \)

Hint:

For capacitors, the force between plates can be derived from the electric field and the charge on the plates. Use the relationship between electric field and charge to find the force.

Answer 1

The Correct Option is B

Step 1: Understanding the relationship.
The force on each plate of a parallel plate capacitor is given by the formula: \[ F = \frac{Q^2}{2A\epsilon_0} \] where \( A \) is the area of the plates and \( \epsilon_0 \) is the permittivity of free space. The electric field between the plates is \( E = \frac{Q}{A\epsilon_0} \). By substituting this expression for \( E \) into the force equation, we get: \[ F = \frac{Q^2}{2A\epsilon_0} = \frac{QE}{2} \] Step 2: Conclusion.
Thus, the force on each plate is \( \frac{QE}{2} \), corresponding to option (B).