Parallel-plate capacitors: The electric field between square the plates of a parallel-plate capacitor has magnitude E. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The magnitude of the electric field between the plates is now equal to

Question

Quizplus · Accepted Answer

The electric field between the plates is given by $E = V/d$, where $V$ is the potential difference between the plates and $d$ is the separation of the plates. Since the voltage is constant, when the separation is doubled, the electric field strength is halved. Thus, the magnitude of the electric field between the plates is now $E/2$. Therefore, the answer is D.

Parallel-Plate Capacitors: the Electric Field Between Square the Plates of a Parallel-Plate