The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart.What happens to the magnitude of the $\vec { E }$ field between the plates during this process?

Question

Quizplus · Accepted Answer

As the plates are pulled apart, the distance between them increases, resulting in a decrease in capacitance. Since the voltage across the plates is held constant, the charge on the plates remains the same. According to the equation $C=\dfrac{Q}{V}$, a decrease in capacitance results in an increase in electric field magnitude for a given charge on the plates. Therefore, the magnitude of the electric field between the plates increases as they are pulled apart.

The Plates of a Parallel-Plate Capacitor Are Maintained with Constant E⃗\vec { E }E

The Plates of a Parallel-Plate Capacitor Are Maintained with Constant $\vec { E }$