One plate of a parallel-plate capacitor with a capacitance of 0.1 F is given a charge of 1 C , while the other plate is given a charge of 2 C . Determine the potential difference developed between the plates.

Step-by-step solution

Let the charges on the two plates be Q₁ and Q₂ . The charge appearing on the inner surfaces of the plates is given by q = |Q₁ - Q₂| 2 . Substituting the given values, Q₁ = 1 C and Q₂ = 2 C : q = |1 - 2| 2 = 0.5 C The potential difference V between the plates is given by V = q C . Substituting q = 0.5 C and C = 0.1 F : V = 0.5 0.1 = 5 V