Charge Q is distributed non-uniformly over a | vedclass

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(C) $1$. The potential at a point $P$ on the axis of a ring at distance $x$ from the center is given by $V = \frac{KQ}{\sqrt{R^2 + x^2}}$. Given $x =

Vedclass · Accepted Answer

Magnitude of electric field at $P$ must be equal to $\frac{3KQ}{10\sqrt{10}R^2}$

Vedclass · Answer

(C) $1$. The potential at a point $P$ on the axis of a ring at distance $x$ from the center is given by $V = \frac{KQ}{\sqrt{R^2 + x^2}}$. Given $x = 3R$,$V = \frac{KQ}{\sqrt{R^2 + (3R)^2}} = \frac{KQ}{\sqrt{10}R}$. Thus,option $A$ is correct.$2$. For a uniformly charged ring,the electric field is $E = \frac{KQx}{(R^2 + x^2)^{3/2}}$. At $x = 3R$,$E = \frac{KQ(3R)}{(R^2 + 9R^2)^{3/2}} = \frac{3KQR}{(10R^2)^{3/2}} = \frac{3KQ}{10\sqrt{10}R^2}$.$3$. Since the charge is distributed non-uniformly,the electric field at $P$ will not necessarily be the same as that of a uniformly charged ring. The field magnitude can vary depending on the distribution. Therefore,the statement that the field must be equal to $\frac{3KQ}{10\sqrt{10}R^2}$ is incorrect. Options $B$ and $D$ are plausible depending on the distribution,making $C$ the wrong statement.

Charge $Q$ is distributed non-uniformly over a ring of radius $R$. $P$ is a point on the axis of the ring at a distance $3R$ from its centre. Which of the following is a wrong statement?

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