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

Question

Consider a parallel-plate capacitor made up of two conducting plates with dimensions 29 mm times 12 mm. If the separation between the plates is 0.55 mm, what is the capacitance, in pF, between them? If there is 0.46 nC of charged stored on the positive plate, what is the potential, in volts, across the capacitor? What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor? If the separation between the plates doubles, what will the electric field be if the charge is kept constant?

Explanation / Answer

(a) We know that the capacitance of capacitor is given by
C = eoA /d
where A is area of plate and d is the distance between them.
C = (8.85*10-12)*(29*12*10-6) /(0.55*10-3) = 5.599 *10-12 F = 5.599 pF
(b) We know that
q= CV
V = q/C = 0.46*10-9 /(5.599*10-12) = 82.148 Volt
(c) Electric field = Potential difference /distance = V/d = 82.148 / (0.55*10-3) = 149.36*103 N/C
(d) We know that
E= V/d
V = q/C, therefore
E = q/C*d = q /(eoA /d)*d = q /eoA
hence the electric field is independent of the distance so when we change the distance electric field will remain constant.

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