A charge of 20 C is deposited on the positive plate of an isolated parallel-plate capacitor having a capacitance of 10 F . Determine the potential difference developed between the plates.

Step-by-step solution

Let the charge given to the first plate be Q₁ = 20 C and the charge on the second plate be Q₂ = 0 . When charges Q₁ and Q₂ are given to the plates of an isolated capacitor, the charge appearing on the inner facing surfaces of the plates is q = Q₁ - Q₂ 2 . Substituting the given values: q = 20 - 0 2 = 10 C The potential difference developed between the plates depends only on the inner surface charges and is given by V = q C . V = 10 C 10 F = 1 V