Question -
Answer -
Let┬аF┬аbe the force appliedto separate the plates of a parallel plate capacitor by a distance of ╬Фx.Hence, work done by the force to do so =┬аF╬Фx
As a result, the potential energy of thecapacitor increases by an amount given as┬аuA╬Фx.
Where,
u┬а=Energy density
A┬а=Area of each plate
d┬а=Distance between the plates
V┬а=Potential difference across the plates
The work done will be equal to the increase in thepotential energy i.e.,
Electric intensity is given by,
However, capacitance,┬а
Charge on the capacitor is given by,
Q┬а=┬аCV
The physical origin of thefactor,┬а
, in the force formula liesin the fact that just outside the conductor, field is┬аE┬аandinside it is zero. Hence, it is the average value,┬а
, of the field thatcontributes to the force.