Question

An air-gap parallel plate capacitor is fully charged by a battery.
What combination of two measurements will allow someone to calculate the magnitude of the electric field in between the capacitor plates?

• A. The potential difference of the battery and the area of the plates.
• B. The charge on the plates and the distance between the plates.
• C. The charge on the plates and the area of the plates.
• D. The area of the plates and the distance between the plates.
• E. More than two measurements are needed to calculate the electric field in between the capacitor plates.

Answer 1

The correct option is C The charge on the plates and the area of the plates.

Capacitance of the parallel plate capacitor $C=\frac{A{ϵ}_o}{d}$

Using $Q=CV$ $⟹V=\frac{Q}{C}$

Electric field between the plates $E=\frac{V}{d}$

$\therefore$ $E=\frac{Q}{Cd}=\frac{Q}{\frac{A{ϵ}_o}{d}\times d}=\frac{Q}{A{ϵ}_o}$

Thus option C is correct.