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Question
A parallel-plate capacitor with the plate area 100 cm2 and the separation between the plates 1⋅0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.
Solution
Area of the plates of the capacitor, A = 100 cm2 = `10^-2 "m"^2`
Separation between the plates, d = 1 cm = `10^-2 "m"`
Emf of battery, V = 24 V
Therefore,
Capacitance , `C = (∈_0A)/d = ((8.85 xx 10^-12) xx (10^-2))/(10^-2) = 8.85 xx 10^-12 V`
Energy stored in the capacitor,
`E = 1/2 CV^2 = 1/2 xx (8.85 xx 10^-12) xx (24)^2`
= `2548.8 xx 10^-12 "J"`
Force of attraction between the plates, `F = E/d = (2548.8 xx 10^-12)/(10^-2) = 2548.8 xx 10^-10 "N"`
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