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My idea is placing charged conducting plates in such a way that they won't see each others' surfaces unlikely to the typical design of parallel plates. If they are placed like this, would be the force that one plate exerting to other same as in the typical design? If we say the charges on the plates are q1 and q2 how could be the force calculated? Since the electric field is perpendicular to surface of a conductor would this placement affect the electrostatic force and field?

Here is how will the plates are placed:

Charged plates sketch

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If the two plates are made of conducting material, there is nothing preventing charges from flowing as close as possible to each other, which, in this case, means toward the edge of each plate closest to the other, right next to the insulating layer.

If we now suppose the layer to be thin (dimension $d$) with respect to the plates' sides $L = 10\; cm$), and also that the plates are thin in the vertical direction, then charge distribution will be essentially linear, with charge density $\lambda = Q/l$ (Q the total charge on each plate). The electric field produced by this configuration, neglecting edge effects (hence the assumption $d\ll l$) is $E = 2\lambda/d$ in a direction perpendicular to the insulating layer axis.

Hence the force per unit length $f = 2\lambda^2/d$, and the total force $$ F = 2 \frac{\lambda^2 l}{d} = 2 \frac{Q^2}{d\; l} $$ is attractive and directed along the line joining the centers of the two plates, for obvious reasons of symmetry.

If instead the two plates are made of insulating material, carrying constant surface charge density $\sigma = Q/l^2$, the force, again directed perpendicular to the insulating layer major axis, can be recovered by means of the usual quadruple integral. Any CAS (=computer algebra system) can do that.

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  • $\begingroup$ Thanks for the answer. It helped me a lot but can you tell me which formulas did you use to calculate the electric field and force? $\endgroup$ – user57144 Aug 14 '14 at 11:14
  • $\begingroup$ @FarstarThe electric field is that of a long wire, see any textbook, or here, farside.ph.utexas.edu/teaching/302l/lectures/node26.html. If E is the field and $\lambda$ the charge per unit length, then the force per unit length is $E\lambda$ and the total force is $E\lambda l$. $\endgroup$ – MariusMatutiae Aug 14 '14 at 11:22
  • $\begingroup$ I looked at the lecture you sent me. I think for this question, the Gauss' law should be modified as E(d)dl=λl/ϵ0 because we need the electric flux for one direction not in every direction. So I found E(d)=λ/dϵ0. Am I right or where am I wrong? $\endgroup$ – user57144 Aug 14 '14 at 11:58
  • $\begingroup$ @FarStar The field cannot be constrained to expand in less than three directions; thus we need to apply Gauss's law to the surface of a cylinder, just like the reference does. As for the $\epsilon_0$,I am using cgs units, that's the only difference. $\endgroup$ – MariusMatutiae Aug 14 '14 at 12:16
  • $\begingroup$ Then what is the value of ϵ0 in cgs units? Is λ/dϵ0 the same as 2λ/d when we use ϵ0 in cgs units? $\endgroup$ – user57144 Aug 14 '14 at 12:22

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