Question

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density \( +\sigma \) and \( -\sigma \). The force experienced by a point charge \( +q \) placed at the mid point between the plates will be:

• A. \( \frac{3q\sigma}{4 \epsilon_0} \)
• B. \( \frac{3q\sigma}{2 \epsilon_0} \)
• C. \( \frac{3q\sigma}{4 \epsilon_0} \)
• D. \( \frac{q\sigma}{2 \epsilon_0} \)

Hint:

When calculating the electric field due to uniformly charged infinite plates, use the formula \( E = \frac{\sigma}{2\epsilon_0} \) for each plate and then sum the fields considering the direction.

Answer 1

The Correct Option is B

Let the charge distribution on the two plates be \( \sigma \) and \( -\sigma \), with the point charge \( q \) placed at the midpoint between the plates.
The electric field due to each plate at the midpoint is as follows:  
For Plate 1, the electric field is \( \frac{\sigma}{2 \epsilon_0} \) directed away from the plate, and for Plate 2, the electric field is \( \frac{\sigma}{2 \epsilon_0} \) directed towards the plate. 
Thus, the net electric field experienced by the charge \( q \) is: \[ E_{\text{net}} = \frac{3\sigma}{2 \epsilon_0} \] Now, the force on the charge \( q \) is given by: \[ F = qE = q \times \frac{3\sigma}{2 \epsilon_0} = \frac{3q\sigma}{2 \epsilon_0} \]