Question

The plates of a parallel-plate capacitor are 3.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuum. The electric field between the plates has a magnitude of 4.00×106 V/m .Part A: What is the potential difference between the plates?Part B:What is the area of each plate?Part C:What is the capacitance?

Answer 1

Answer: a) 14*10^3 V; b)2.4 *10^-3 m^2; c)6.07*10^-12 F

Explanation: In order to explain this problem we have to consider, the following expressions:

V=E*d the difference of potencial in a capacitor is equal to the electric field multiply the separacion between the plates then,

V= 4*10^6*3.5*10^-3=14*10^3 V

We also know that the electric field in a parallel plates capacitor is given by:

E=σ/εo where σ is Q/A ( charge/area)

E=Q/(A*εo)=85*10^-9/(A*8.85*10^-12) then

A=85*10^-9/(8.85*10^-12*4*10^6)=2.4 *10^-3 m^2

Finally, the capacitance is given by:

C= εo*A/d

C=8.85*10^-12*2.4*10^-3/3.5*10^-3=6.07*10^-12 F