capacitor

suppose kar lo q charge on leftmost surface.. conserve charge and put q₁-q charge on surface b..

then by induction charge on surface c = -(q₁-q)

again conserve charge for plate 2 to find charge on surface d..

then again induction induces charge on surface e

conserve charge for plate 3 to find charge on surface f..

now use the surface charge distribution INSIDE any of the plates .. it will come in terms of q₁, q₂, q₃ and q...

Then equate the field to 0 because there can be no field inside a conductor..

that gives u q in terms of q₁, q₂ and q₃..

thus u come to know the amount of charge on each surface! :)

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