A parallel-plate capacitor has (one) plate area of 26 cm $2$ and a separation distance between the two plates of 1.1 mm. The potential across the plates is 12 V. The electric field between the plates is
A) 170 N/C.
B) 11,000 N/C.
C) 1700 N/C.
D) 5600 N/C.

Question

Quizplus · Accepted Answer

The electric field (E) between the plates of a capacitor is given by the formula $E = \frac{V}{d}$, where $V$ is the potential difference across the plates and $d$ is the separation distance between the plates. Here, $V = 12$ V and $d = 1.1$ mm = $0.0011$ m. Thus, $E = \frac{12}{0.0011} = 10909.09$ N/C, which is approximately $11,000$ N/C.

A Parallel-Plate Capacitor Has (One) Plate Area of 26 Cm 222

A Parallel-Plate Capacitor Has (One) Plate Area of 26 Cm $2$