A capacitor of capacitance `5muF` is charged to a potential difference 100V and then diconnected from the powerr supply. The minimum work needed to pull the plates of the capacitor apart so that the distance between them doubles is (in mJ).

Explanation

When capacitor is charged its capacitance is C and the potential difference applied is V. Hence the stored charge will be Z = CV. The amount of energy stored in the capacitor:
`U_(1)=(1)/(2)CV^(2)" "…(i)`
After we disconnect the capacitor from the battery, its charge Q = CV remains constant. When separation between the plates becomes twice its initial value then the capacitance must get reduced to half its initial value. Energy stored in the capacitor, after the separation is made double, can be written as follows:
`U_(2)=(Q^(2))/(2(C_("new")))=(C^(2)V^(2))/(2xx(C )/(2))`
`rArr U_(2)=CV^(2)" " ...(ii)`
From energy conservation we can write the following equation:
`U_(2)=U_(1)+W_("ext")`
`rArr W_("ext")=U_(2)-U_(1)=CV^(2)-(1)/(2)CV^(2)=(1)/(2)CV^(2)`