Answer:
[tex]E=\dfrac{KQx}{(x^2+a^2)^{\frac{3}{2}}}[/tex]
Step-by-step explanation:
Given that
Radius of ring=a
Distance of point P from center=x
Total charge=Q
We know that electric field is given by
[tex]E=\dfrac{KQcos\theta }{r^3}[/tex]
Here electric field in two direction will be cancel out and only in one direction electric field will exits.
[tex]cos\theta =\dfrac{x}{r}[/tex]
[tex]r^2=x^2+a^2[/tex]
[tex]E=\dfrac{KQx}{(x^2+a^2)^{\frac{3}{2}}}[/tex]
