Question

Two large, thin, metal plates of 0.5mx0.5m face each other. They are spaced 2cm apart and have equal but opposite charges on their inner surfaces.

a) if the magnitude E of the electric field between the plates is 700 N/C what is the magnitude of the charge on each plate?

b) Integrate E across the gap between the plates to find the potential difference V and hence the capacitance C.

c) Recalculate C using the parallel plate capacitator formula instead.

d) the energy density of the field between the plates?

e) the total energy of the field between the plates?

f) use C and V to calculate the energy stored in the capacitator to compare to the result of e

If a very detailed explanation could be given that'd be great because im super stuck!

Answer 1

a) The magnitude of the Electric Field between the two plates is given by E = Q/E0.A.....(1); where Q is the charge on each plate, E0 is the permittivity of free space = 8.85 X 10^-12 F/m and A is the area of cross section of each plate = 0.25 m^2(given)

E = 700 N/C(given)

From (1), we have 700 = Q/8.85 X 10^-12 X 0.25

Solving, we get Q = 1.54875 X 10^-9 C or 1.54875 nC Ans.

b) Potential Difference(V) =

Distance between the plates (d) = 2 cm = 0.02 m

=

= E (0.02-0)

= 700 X 0.02

= 14 V Ans

Capacitance(C) = Q/V

= 1.54875 X 10^- 9/ 14

= 1.10625 X 10^-10 F

= 0.110625 nF Ans.

c) C = E0.A/d

= 8.85 X 10^-12. 0.25/ 0.02

= 0.110625 nF Ans.

d) Energy Density = 1/2 E0. E^2

= 1/2. 8.85 X 10^-12 X 700^2

= 2.16825 X 10^-6 J/m^3

= 2.16825 uJ/m^3 Ans.

e) Total Energy Stored = Energy Density X Volume

= 2.16825 X A.d

= 2.16825 X 0.25 X 0.02

= 0.01084125 uJ Ans.

f) Energy stored in capacitor = 1/2 C.V^2

= 1/2 X 0.110625 X 10^-9 X 14^2

= 1.084125 X 10^-8 J

= 0.01084125 uJ Ans, which is the same as obtained in e).