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Unit of Force

Mechanical force on unit area of a charged conductor

Consider a charged conductor of any shape situated in a medium of dielectric constant k (Fig. 13.7). Let dS be a small area of the surface of the conductor, if a is the surface density of charge over the area dS, the electric intensity at a point just outside the conductor is given by

E = ^ ...(13.17)



andit acts in a direction perpendicular to the surface of the conductor. This intensity can be considered to be the resultant of two intensities; (i) The intensity El due to the charge on the area dS and (ii) The intensity E2 due to the charge on the remaining surface of the conductor. Thus we can write

E = Ei+ E2=^ .-(13.18)


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The intensity Ex is due to the charge on dS and hence it is directed on both sides of the area dS as shown in Fig. (13.7). However the intensity E2 due to the remaining charge on the conductor is directed only in the outward direction.


The resultant intensity at the internal point B is El - E2. But the intensity at B must be zero, as there is no charge inside the conductor.

El- E2 =0 E j = E2 Therefore from Eq. (13.18), we get

Ex = E2 = ~ ...(13.19)


It can thus be seen that the intensity just outside the area dS of

a

the charged conductor is E = due to the charge on the area it a

self and E2 = £ due to the remaining charge on the conductor, so a

that the resultant intensity is E- 7. &0k

The charge on the area dS is thus situated in the electric field of intensity E2 produced at the area dS by the rest of the charge on the conductor. Therefore a force F will act on the area dS given by

F = (charge on dS) x (intensity due to remaining charge) ' a ^

= (adS) x E2 = adS x I ° J a2

= — dS ...(13.20)


Therefore the force if) acting on unit area of the charged conductor is given by

/ = ^ ...(13.21)

This expression gives the force per unit area, in terms of the surface density of charge.

CT

Now from eq. (13.17), E - - /. o = eokE

e0k

Substituting in Eq. (13.21) we get

(e0kE)2 1

f = ^T = 2 £°kg! -(13-22)


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This relation gives the force per unit area in terms of the electric intensity.

Example 1

A charge of -Js.S5 M-C is uniformly distributed on the surface of a metallic object of area 3m2, placed in air. Calculate the mechanical force acting on it. ( £0= 8.85 x 10"12 C2/Nm2). Solution : Given : Q = V8.85 M-C =^8.85 x 10~6 C, A = 3 m2

£0 = 8.85 x 10-12 C2/Nm2, k = 1 (for air)

a2 Q2 ( _Q) Force per unit area, f="——-= —^-- • ct_t

F 2s0k 2A e0k I AJ

Q2

Force on metal surface, F = —-- x A

2A s0k

_ 8.85xl0~12x3

2x9x8.85xl0-12 xl

= 0.1667 N ...Ans