The electric field in a parallel plate capacitor has magnitude 1.40 x 104 V/m. What is the surface charge density (LaTeX: / sigma σ, in C/m2) on the positive plate? (LaTeX: / epsilon_0 ϵ 0 = 8.854 x 10-12 F/m)

Question

Genevieve

Chemistry

6 August, 18:50

The electric field in a parallel plate capacitor has magnitude 1.40 x 104 V/m. What is the surface charge density (LaTeX: / sigma σ, in C/m2) on the positive plate? (LaTeX: / epsilon_0 ϵ 0 = 8.854 x 10-12 F/m)

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Answers (1)

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Jacqueline Wang · Answer 1

In the parallel plate capacitor,

q = e0EA, q is the charge on the plate, E is electric field intensity, A is the area of each plate, e0 = 8.854 x 10^-12F/m.

The charge density is q/A (charge per unit area)

q/A = e0 x E, from the equation above.

Therefore, q/A = 1.40 x 10^4V/m x 8.854 x 10^-12F/m

q/A = 12.3956 x 10^-8 = 1.23956 x 10^-7C/m2

Explanation:

A parallel plate capacitor consist of two plates facing eachother with separation d in between them. One of the plate is positively charged and the other is negatively charged. There is a dielectric material in between the plates. The charges on the plate is given as: q = e0EA