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

                                                           figure (a)

                       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.