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)
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)
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
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.
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