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Question

# The surfaces of these conductor plates are given charges as shown and are separated by very small distances compared to their surface area. What will be the charge on the surface 'e'?

A
+5C
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B
-2C
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C
-5C
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D
+2C
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Solution

## The correct option is C -5COk! If you went through the hint and couldn’t exploit the Gauss Law, it’s Okay. This one example will help you develop the thought process required to do that. The first thing we would do is consider a cuboidal Gaussian surface through the material of the conductor (Why?-you’ll get to know in a minute) Something like this! I have just re-drawn the conductor plates showing their thickness. The blue rectangle is the side profile of the cuboidal Gaussian surface that I chose. Now the electric field inside the conductor surface is zero so the flux through that part of the cuboid would be zero and the part of the cuboid outside would also have no flux (think why?). so, the net flux through the Gaussian surface is zero, hence from Gauss’s law the net charge enclosed must be zero! That is if a charge q comes on side b then charge on side c must be –q. That makes charge on side a equal to 3-q (from law conservation of charge) and on side d is 5+q That makes charge on side e equal to –(5+q) from the same logic. The charge on f becomes -2+(5+q)=3+q Now if I consider a point P ‘inside’ the third plate as shown, the net electric field there must be zero! Electric field at P is due to all the plates =>Ea+Eb+Ec+Ed+Ee+Ef=0 ⇒3+q2Aε0+q2Aε0−q2Aε0+5+q2Aε0−5+q2Aε0+−2+5+q2Aε0=0 (considering A as the surface area of the conductor and using the formula δ2ε0 (as the distances b/w the plates is too small compared to their surface areas) ⇒6+2q=0⇒q=−3 So the final charge distribution would be

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