Electrical Engineering ⇒ Topic : Force on Charged Plates
Force on Charged Plates
Consider two parallel conducting plates x metres apart and carrying constant charges of +Q and -Q coulombs respectively as shown in Fig. (a). Let the force of attraction between the two plates be F newtons. If one of the plates is moved away from the other by a small distance dx, then work done is
Work done = F x dx joules .......(1)
Since the charges on the plates remain constant, no electrical energy can enter or leave the system during the movement dx.
Work done = Change in stored energy
Since the separation of the plates has increased, the capacitance will decrease by dC. The final capacitance is, therefore, (C - dC).
This represents the force between the plates of a parallel-plate capacitor charged to a p.d. of volts. The negative sign shows that it is a force of attraction.
Note. The force of attraction between charged plates may be utilised as a means of measuring potential difference. An instrument of this kind is known as an electrostatic voltmeter.
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