Consider a parallel-plate capacitor made up of two conducting plates with dimens

ID: 1653432 • Letter: C

Question

Consider a parallel-plate capacitor made up of two conducting plates with dimensions 42mm x 47mm

(1)If the separation between the plates is 0.65mm, what is the capacitance, in pF, between them?

(2)If there is 0.13 nC of charged stored on the positive plate, what is the potential, in volts, across the capacitor?

(3)what is the magnitude of the electric field, in newtons per coulomb, inside the capacitor?

(4)If the separation between the plates doubles, what will the electric field be if the charge is kept constant?

(1) C = e0 A / d

C = (8.854 x 10^-12) (42 x 10^-3 x 47 x 10^-3) / (0.65 x 10^-3)

C = 26.9 x 10^-12 F or 26.9 pF

(2) Q = C V

0.13 x 10^-9 = (26.9 x 10^-12) V

V = 4.83 Volt

(3) E = V / d = (4.83) / (0.65 x 10^-3)

E = 7431 N/C

(4) elctric field = q / e0 A

q, A are not changing hence field will remain same.

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