Question

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

 

Answer 1

Let the total charge of the ring be Q.
Radius of the ring = R
The electric field at distance x from the centre of ring, 

\[E = \frac{Qx}{4\pi \epsilon_0 \left( R^2 + x^2 \right)^{3/2}}         .  .  . (1)\] 

For maximum value of electric field,

\[\frac{dE}{dx} = 0\]

From equation (1), 

\[\frac{dE}{dx} = \frac{Q}{4\pi \epsilon_0}\left[ 1( R^2 + x^2 )^{- 3/2} - x\frac{3}{2}( R^2 + x^2 )^{- 5/2} 2x \right]=0\]

\[\Rightarrow  R^2  +  x^2  - 3 x^2  = 0\] 

\[ \Rightarrow 3 x^2  =  R^2 \] 

\[ \Rightarrow x = \frac{R}{\sqrt{2}}\]