The capacitance between adjacent plates in a three-plate arrangement is 50 ext nF . A charge of 1.0 ext C is placed on the middle plate. Calculate the potential difference developed between the upper and the middle plates.

Step-by-step solution

Let the charge given to the middle plate be Q = 1.0 C . By symmetry, the charge Q distributes equally on the two faces of the middle plate. Therefore, the charge on the upper face of the middle plate is q = Q 2 = 0.5 C . The upper plate and the middle plate act as a parallel plate capacitor with capacitance C = 50 nF . The potential difference V between the upper and middle plates depends on the charge on their facing surfaces. Using the relation V = q C : V = 0.5 10⁻⁶ 50 10⁻⁹ = 500 10⁻⁹ 50 10⁻⁹ = 10 V