Answer :


Given:
Charge of the ring: Q
Radius of the ring :R
Let P be the point where electric field is found.
Distance between center of the ring and P is x.

Formula used:
We know that electric field at any point on the axis at a distance x from the center is:
Where k is a constant and k= =9× 109 Nm2C-2. Q is the charge of the ring, x is the distance between center of the ring and point P and R is the radius of the ring.
Now for the electric field to be maximum, we use the maxima property:
Taking derivative of E w.r.t x,






Hence Electric field is maximum at on the axis.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Five charges, q ePhysics - Exemplar

Fig. 1.10 represePhysics - Exemplar

The electric forcHC Verma - Concepts of Physics Part 2

A particle of masHC Verma - Concepts of Physics Part 2

The bob of a simpHC Verma - Concepts of Physics Part 2

A positive chargeHC Verma - Concepts of Physics Part 2

A rod of length LHC Verma - Concepts of Physics Part 2

Consider a uniforHC Verma - Concepts of Physics Part 2