Force of Attraction Between Oppositely-charged Plates
In the figure given below, shown two parallel conducting plates A and B carrying constant charges of + Q and − Q coulombs respectively. Let the force of attraction between the two be F newtons. If one of the plates is pulled apart by distance dx, then work done is
= F × dx joules ………………………………………….(i)
Since the plate charges remain constant, no electrical energy comes into the arrangement during the movement dx.
∴ Work done = change in stored energy
Initial stored energy
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If capacitance becomes (C − dC) due to the movement dx, then Final stored energy
∴ Change in stored energy
Equating Eq. (i) and (ii), we have
This represents the force between the plates of a parallel-plate capacitor charged to a potential difference of V volts. The negative sign shows that it is a force of attraction.
Read article – parallel-plate capacitor
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