Assertion: For a non-uniformly charged thin c | vedclass

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(D) For a thin circular ring with a non-uniform charge distribution and a net charge of $0$,the electric potential $V$ at any point $P$ on the axis is

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If Assertion is incorrect but Reason is correct.

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(D) For a thin circular ring with a non-uniform charge distribution and a net charge of $0$,the electric potential $V$ at any point $P$ on the axis is given by $V = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r}$. Since the net charge $\int dq = 0$ and all points on the ring are at the same distance $r$ from any point on the axis,the potential $V$ is indeed $0$ at every point on the axis.However,the electric field $\vec{E}$ is the negative gradient of the potential,$\vec{E} = -
abla V$. While the potential is zero everywhere on the axis,it does not mean the potential is constant in the neighborhood of the axis. In fact,the electric field on the axis is generally non-zero and directed perpendicular to the axis. Thus,the Assertion is incorrect,and the Reason is correct.

Assertion: For a non-uniformly charged thin circular ring with net charge $0$,the electric field at any point on the axis of the ring is zero.
Reason: For a non-uniformly charged thin circular ring with net charge $0$,the electric potential at each point on the axis of the ring is zero.

Similar Questions

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As shown in the figure,at what distance in $cm$ from point $A$ will the electric field be zero?

The given diagram shows two semi-infinite lines of charge having equal (in magnitude) linear charge density but with opposite signs. The electric field at any point on the $x$-axis for $(x > 0)$ is along the unit vector:

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Assertion: For a non-uniformly charged thin circular ring with net charge $0$,the electric field at any point on the axis of the ring is zero.Reason: For a non-uniformly charged thin circular ring with net charge $0$,the electric potential at each point on the axis of the ring is zero.