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Electricity

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4 Jul 2009 19:22:00 IST

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231

Gauss's law

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Two conducting paltes X and Y ech having large surface area A (on one side) are placed parallel to each other as shown in figure. The platr X is given a charge Q whereas the other is nuetral. Find A)the surface charge density at the inner surface of the plate X. B) the electric field at a pont to the left of the plates, C) the electric field at a point in between the plates asnd d) ther electric field at a point to the right of the plates.

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Debotosh Chatterjee

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4 Jul 2009 22:07:23 IST

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where is the figure????????

VARUN RAJ

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5 Jul 2009 20:51:49 IST

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i am assuming that the area of both the sides are equal assuming

charge on outer side of the x to be q1 and on the inner side of q-q1

charge on the inner side of y is -(q-q1) because if we draw a gaussian surface passing thru the conductore the electric field inside should be -(coz of conduction properties)

charge on outer side=q-q1

the electric field at any pnt x or y =0 as it is inside the conductor

so q-q1=q solving we get q1=q2

so now we can get the all the answers easily

M. Fuaad Khan

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7 Jul 2009 00:32:51 IST

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Since charge q is distributedon metallic plate, so the charge distributes as q/2 & q/2

=> = q/2A

Now E due to metallic face = / ₀

By induction of charge -q/2 E due to adjacent face on other metallic plate

E = - / ₀

=> E on either side of of plate = / ₀- / ₀ = 0

E between plates = / ₀+ / ₀ = 2 / ₀

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