Question

A parallel plate capacitor is constructed with circular plates of radius 0.750 cm and plate separation 0.0500 mm. If the capacitor is connected across a 37.2 V source, find:

a) the capacitance

b) the surface charge on each plate

c) The energy stored in the capacitor

d) the electric field between the plates

e) the energy density between the plates

Answer 1

C = e0 A/ d= 8.854 * 10^-12 * ( pi * ( 0.75 * 10^-2) ^2 ) / ( 0.05 * 10^-3)

= 31.29 * 10^-12 F

2) Q = CV

= 31.29 * 10^-12 * 37.2

= 1.164*10^-9 C

3)

E = 1/2 CV^2

= 1/2 * 31.29 * 10^-12 * ( 37.2)^2

= 2.165 *10^-8 J

4) E = V d

= 37.2 * 0.05 * 10^-3

= 1.86 *10^-3 N/C

5) Energy density = Energy / Area

= 2.165 * 10^-8 / (pi * ( 0.75 * 10^-2) ^2 )

= 1.225 *10^-4 J/m^2

Answer 2

Answer

C = e0 A/ d= 8.854 * 10^-12 * ( pi * ( 0.75 * 10^-2) ^2 ) / ( 0.05 * 10^-3)

= 31.29 * 10^-12 F

2) Q = CV

= 31.29 * 10^-12 * 37.2

= 1.164*10^-9 C

3)

E = 1/2 CV^2

= 1/2 * 31.29 * 10^-12 * ( 37.2)^2

= 2.165 *10^-8 J

4) E = V d

= 37.2 * 0.05 * 10^-3

= 1.86 *10^-3 N/C

5) Energy density = Energy / Area

= 2.165 * 10^-8 / (pi * ( 0.75 * 10^-2) ^2 )

= 1.225 *10^-4 J/m^2

Answer 3