Electrical Engineering ⇒ Topic : Force of Attraction between Oppositely Charged Plates

FORCE OF ATTRACTION BETWEEN OPPOSITELY CHARGED PLATES

Figure (a) shows two parallel plates separated by a distance x metre and carrying charge +Q and -Q coulombs, respectively. Let the force of attraction between the two plates be F newton. Let one of the plates be pulled apart by a distance dx, the work done is given by

                                                                W = F x dx joule

During the movement dx, no electrical energy comes into play as the plate charge remains constant.

                                      Work done = Change in stored energy                                                                     ................... (1)

                                       ............. (1)

                                                          .............. (2)

Thus, from Eqs. (1) and (2), we get

Consider two parallel conducting plates x metres apart and carrying constant charges of +Q and -Q coulombs respectively as shown in Figure (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)

                                                            FIGURE (A)

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

                         Initial stored energy =

Since the separation of the plates has increased, the capacitance will decrease by dC. The final capacitance is, therefore, (C - dC).

                         Final stored energy =

Since dC is small compared to C, (dC)² can be neglected compared to C²

^{Final stored energy}

^{Change in stored enegry =}              ........   (2)

Equating eqs. (1) and (2), we get,

                                                                         ............ (3)   (V =Q/C)

Substituting the value of dC/dx in eq. (3), we get,

                                           ...........in a medium

                                                 .........in air

This represents the force between the plates of a parallel-plate capacitor charged to a p.d. of V 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.