(a) Write two important characteristics of equipotential surfaces.

(b) A thin circular ring of radius r is charged uniformly so that its linear charge density becomes \lambda. Derive an expression for the electric field at a point P at a distance x from it along the axis of the ring. Hence, prove that at large distances (x >> r), the ring behaves as a point charge.

 

 
 
 
 
 

Answers (1)

a) For equipotential surfaces
(i) Potential has the same value at all points on the surface.
(ii)The electric field is normal to the equipotential surface at all points

b)

At P the electric field due to dq

\\dE=\frac{kdq}{x^2+r^2}\\dE_{x}=dEcos\theta=dE\frac{x}{\sqrt{x^2+r^2}}

The component, perpendicular to the axis gets cancelled by the field due to another elemental charge symmetrically located on the other side of the axis.

\\E=\int dE\cos\theta

\\=\int k\frac{dq}{x^2+r^2}\frac{x}{\sqrt{x^2+r^2}}\\\\=\frac{kx}{(x^2+r^2)^\frac{3}{2}}\int \lambda dl\\\\=\frac{1}{4\pi\epsilon_0}\frac{x\lambda\times2\pi r}{(x^2+r^2)^{\frac{3}{2}}}

\\=\frac{Qx}{4\pi\epsilon_0(x^2+r^2)^{\frac{3}{2}}}

for x>>r

E=\frac{Q}{4\pi\epsilon_0x^2}

This corresponds to the expression for the electric field due to a point charge. Thus at large distances, the ring behaves like a point charge.

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