The figure shows a nonconducting ring which has positive and negative ...
Explanation:
To prove that the potential at all points on the axis will be zero, we need to consider the symmetry of the charge distribution on the nonconducting ring.
Symmetry:
Since the charge distribution on the ring is nonuniform but the total charge is zero, it implies that the positive and negative charges are distributed in such a way that the symmetry is maintained.
Electric Field:
The electric field at any point on the axis due to a small element of charge on the ring can be calculated using Coulomb's Law. However, due to the symmetry of the charge distribution, the electric fields due to each small element of charge will cancel out each other along the axis.
Proof:
Consider a small element of charge on the ring at an angle θ. The electric field due to this small element can be written as:
dE = k * dq / r^2
where k is the electrostatic constant, dq is the charge of the small element, and r is the distance between the small element and the point on the axis.
Since the charge distribution is symmetric, the magnitude of the electric field due to a positive charge element at θ will be equal to the magnitude of the electric field due to a negative charge element at θ+π. However, the direction of the electric field due to the positive charge element will be opposite to the direction of the electric field due to the negative charge element.
When we sum up the electric fields due to all the small charge elements on the ring, the electric fields along the axis due to the positive and negative charge elements at θ and θ+π will cancel each other out.
This cancellation of electric fields will occur for every pair of positive and negative charge elements on the ring. Therefore, the net electric field at all points on the axis will be zero.
Conclusion:
Since the electric field at all points on the axis is zero, the potential at all points on the axis will also be zero. Hence, option 'A' is correct.