Question

How much energy is stored by the electric field between two square plates, 6.1 cm on a side?

Answer 1

Answer:

Energy in the capacitor is

E=0.5 CV^2

The capacitance can be approximately defined as

C=e A/d

Since voltage is the ration between charge Q stored and Capacitance C we have

V= Q/C

Substituting Q/C into energy equation we have

E=0.5 C(Q/C)^2

E= Q^2/(2 C)

E= Q^2 d/(2 e A)

e= dielectric constant=8.854 E−12 F/m (or C^2/(N m^2))

E= (520 E-6)^2 x 1.5 E-3/(2 x 8.854 E−12 x (6.4 E-2)^2)

E=5,810 J

PLZ FOLLOW ME

Explanation:

Answer 2

Answer:

Energy in the capacitor is

E=0.5 CV^2

The capacitance can be approximately defined as

C=e A/d

Since voltage is the ration between charge Q stored and Capacitance C we have

V= Q/C

Substituting Q/C into energy equation we have

E=0.5 C(Q/C)^2

E= Q^2/(2 C)

E= Q^2 d/(2 e A)

e= dielectric constant=8.854 E−12 F/m (or C^2/(N m^2))

E= (520 E-6)^2 x 1.5 E-3/(2 x 8.854 E−12 x (6.4 E-2)^2)

E=5,810 J

PLZ FOLLOW ME

Explanation: