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 E_{l} due to the charge on the area dS and (ii) The intensity E_{2} due to the charge on the remaining surface of the conductor. Thus we can write

^{E} = ^{E}i^{+ E}2^{=}^ .-(13.18)

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The intensity E_{x} 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 E_{2} due to the remaining charge on the conductor is directed only in the outward direction.

The resultant intensity at the internal point B is E_{l} - E_{2}. But the intensity at B must be zero, as there is no charge inside the conductor.

E_{l}- E_{2} =0 E j = E_{2 }Therefore from Eq. (13.18), we get

E_{x} = E_{2} = ~ ...(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 E_{2} = £ due to the remaining charge on the conductor, so a

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

The charge on the area dS is thus situated in the electric field of intensity E_{2} 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 E_{2} = adS x I ° J a^{2}

= — 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 = e_{o}kE

e_{0}k

Substituting in Eq. (13.21) we get

(e_{0}kE)^{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 3m^{2}, placed in air. Calculate the mechanical force acting on it. ( £_{0}= 8.85 x 10"^{12} C^{2}/Nm^{2}). Solution : Given : Q = V8.85 M-C =^8.85 ^{x} 10~^{6} C, A = 3 m^{2}

£_{0} = 8.85 x 10-^{12} C^{2}/Nm^{2}, k = 1 (for air)

a^{2} Q^{2} ( _Q) Force per unit area, f="——-= —^-- • ^{ct_}t

^{F} 2s_{0}k 2A e_{0}k I ^{A}J

Q^{2}

Force on metal surface, F = —-- x A

2A s_{0}k

_ 8.85xl0~^{12}x3

2x9x8.85xl0^{-12} xl

= 0.1667 N ...Ans