Cavity

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see, it will depend on the position of 'x'.

case 1: if x < = R/2

E.4Î x² = q/ε0

here, q=0.

so, potential (V=Ed) in this region = 0.

case 2 : if R/2 <x< R

q = find the charge containd 'within' the radius x.

then apply gauss law, find E.

then, V= Ex.

case 3 : if x> R

q= Q.

then similar steps as in case 2.

if it is a conductor

then there will be electric field only x≥R

E4∩x²=Q/ε₀

E=-dV/dx

hmm... point ....

tx sankara..

i actually dunno y ... din think abt it only ...

oops i am also wrong with my reasons

can u correct it sky