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Question

Find the electric potential at the axis of a uniformly charged disc and use potential to find the electric field at same point.

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Solution

Potential due to ring element of radius r and thickness dr is dv=R014πε0σ(2πrdr)r2+x2
V=2σ4πε0[r2+x2]R0=σ2ε0(R2+x2x)
E=dVdx=σ2ε0(2x2R2+x21)=σ2ε0(1xR2+x2)
1034827_1017821_ans_12a74d454b6944cab077f5a1ddf62303.png

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