Doubt from electrostatics


Give solution with detailed explanation.
Q no:5 (multi correct answer type)
Reference:Cengage physics.


It should be A and D.


Can you pls explain why it is A and D

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Electric field due to charges on the plate is only inside the plate as charge appear on outer surface is zero. So electric field outside the plate is E and inside the plate is E+(Q/A€) . Ask if you have any doubt.



See, the arrangement looks more or less as a charged capacitor placed in an electric field.
Its clear and intuitive that the field at R is E only, equal to the external field, so A is correct.

Now, due to the charges that are present on the plates, we will have this sort of distribution of fields :

So here you can see that, the net field towards the left of plate 'a' is : E_{out}+E_b-E_a=E_{out} (as both E_a and E_b have the same magnitude equal to \frac{Q}{2A\epsilon_0} which could be derived using the Gauss' Law. This justifies the fact that the net field at R is E_{out} only.

Similar case is towards the right of the plate 'b' and hence the net field at T is also E_{out}.

However in the region between the two plates : E_{net}=E_{out}+E_a+E_b=E_{out}+\frac{Q}{A\epsilon_0}. And hence, the net field at S will be the same.

Do let me know if you have any doubts.

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Sir my doubt is capacitors are conducting plates (conductors) .So charges rearrange in order to make net electric field inside the conductor zero.In that sense how charge rearrange in capacitors now?(Cause in your solution net electric field inside the plates is not zero) which is not convincing me?

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You should note that, the net electric field inside any conductor is zero, and so will follow here also. The net field inside the rods individually will be zero only, but the regions asked are not inside the rod, but outside it, and hence we can clearly apply the superposition of fields in that region. The reason I pointed out as a capacitor was just to indicate that the net field inside the region between two plates (due to their charges alone) will be \frac{Q}{A\epsilon_0} but the total field overall will be the sum of this one and the external field.

Do let me know if you have any doubts.