Question

A 600 pf capacitor is charged by a 200 v supply. calculate the electrostatic energy stored in it. it is then disconnected from the supply and is connected in parallel to another uncharged 600 pf capacitor. what is the energy stored in the combination?

Answer 1

Answer:

given =  C1 = 600 pF

               V1 = 200V

              C2 = 600 X 10⁻¹²

              V2 = 0

to find = energy stored =?

solution =

                   U1 = 1/2  c1v1²

                   U1 = 1/2 X 600 x 10⁻¹¹²

                   U1 = 12 X 10⁻⁴

 Total charge = Q = q1 + q2

                              =  C1V1 + C2V2

                              =   120000 x 10⁻¹²

                              =  12 x 10⁻⁸

now for capicitance = C = C1 + C2

                                           600 +600

                                           1200 x 10⁻¹²

for total potential =   total charge / total  capacitance

                         V =  12 x 10⁻⁸ / 1200 x 10⁻¹²

                         V = 100

                       U2 = 1/2C2V²

                       U2 = 1/2 X 1200 X 10⁻¹² X 10000

                       U2 = 6 X 10⁻⁴J

loss in energy = U1 -U2

ΔU = 12 X 10⁻⁴ - 6 X 10⁻⁶

6X 10⁻⁴

the energy stored in the combination  6X 10⁻⁴