The distance between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?

Step-by-step solution

For an isolated charged capacitor, the charge Q on the plates remains constant. The capacitance of a parallel-plate capacitor is given by C = ₀ A d . As the distance d between the plates increases, the capacitance C decreases. The potential difference across the capacitor is V = Q C . Since Q is constant and C decreases, the potential difference V increases. The energy stored in the capacitor is U = Q^2 2C . Since C decreases, the energy U increases. The electric field between the plates is E = V d = Q A ₀ . Since