A parallel-plate capacitor has (one) plate area of 7.6 cm $2$ and a separation distance between the two plates of 0.55 mm. There is a nylon dielectric, $k = 3.5$ , between the plates. The potential across the plates is 12 V. The electric field between the plates is
A) 2500 N/C.
B) 3100 N/C.
C) 4300 N/C.
D) 6200 N/C.

Question

Quizplus · Accepted Answer

The electric field (E) between the plates of a capacitor is given by E = V/d, where V is the potential difference and d is the separation distance. Here, V = 12 V and d = 0.55 mm = 0.00055 m. Therefore, E = 12 / 0.00055 = 21818.18 N/C. However, the dielectric reduces the electric field by a factor of k, so E = 21818.18 / 3.5 ≈ 6200 N/C.

A Parallel-Plate Capacitor Has (One) Plate Area of 7 222 And a Separation Distance Between the Two Plates of 0

A Parallel-Plate Capacitor Has (One) Plate Area of 7 $2$ And a Separation Distance Between the Two Plates of 0