Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.

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Solution

The electric field at distance x from the centre of ring is
$E = \frac{Q}{4 π ϵ_{0} (R^{2} + x^{2})^{\frac{3}{2}}}$
For maximum value of eelctric field
$\frac{d E}{d x} = 0$
From equation (i)
$\frac{d E}{d x} = \frac{Q}{4 π ϵ_{0}}$
$[1. (R^{2} + x^{2})^{\frac{3}{2}} - x . \frac{3}{2} (R^{2} + x^{2})^{\frac{1}{2}} .2 x] = 0 o r R^{2} + x^{2} - 3 x^{2} = 0 o r 2 x^{2} = R^{2} ∴ x = \frac{R}{\sqrt{2}}$

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Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.

The electric field at distance x from the centre of ring is E=Q4πϵ0(R2+x2)32 For maximum value of eelctric field dEdx=0 From equation (i) dEdx=Q4πϵ0 [1.(R2+x2)32−x.32(R2+x2)12.2x]=0orR2+x2−3x2=0or2x2=R2∴x=R√2