Question

A 90 pF capacitor and a 360 pF capacitor are both charged to 2.6 kV. They are then disconnected from the voltage source and are connected together, positive plate to positive plate and negative plate to negative plate. Find the energy lost when the connections are made.

________J

Answer 1

initial charge q₁ = 90 X 10^-12 X 2.6 X 10³

= 2.34 X 10^-7 C

q₂ = 360 X 10^-12 X 2.6 X 10³

= 9.36 X 10^-7 C

net total charge = 9.36 X 10^-7 C - 2.34 X 10^-7 C = 7.02 X 10^-7 C

capacitors connected in parallel = 90 + 360 = 450 pF

then voltage = 7.02 X 10^-7 / 450 X 10^-12

= 1560 volts

the energy before is = 0.5 ( 90 X 2600² + 360 X 2600² )

= 1.521 X 10^-3 J

after

= 0.5 X 1560² X 10^-12 X ( 90 + 360 )

= 5.475 X 10^-4 J

so loss is = 1.521 X 10^-3 J - 5.475 X 10^-4 J

= 9.735 X 10^-4

= 973.5 X 10^-6

= 973.5 J

so the the energy lost when the connections are made is 973.5 J