Question

A capacitor with capacitance 5µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?

Options

• 6.25 × 10^-6 J

• 3.75 × 10^-6 J

• 2.16 × 10^-6 J

• 2.55 × 10^-6 J

Answer 1

3.75 × 10^-6 J

Explanation:

U = U_f - U_i

= `"q"/2(1/"C"_"f"-1/"C"_"i")`

= `((5xx10)^2)/2 (1/2-1/5)xx10^6`

= 3.75 × 10^-6 J