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Question

(D) The electric field due to a ring of radius $R$ with charge $-Q$ at a point on its axis at distance $x$ from the center is given by $E_{ring} = \fr

Vedclass · Accepted Answer

$\frac{+Q}{2\sqrt{2}}$

Vedclass · Answer

(D) The electric field due to a ring of radius $R$ with charge $-Q$ at a point on its axis at distance $x$ from the center is given by $E_{ring} = \frac{k Q x}{(R^2 + x^2)^{3/2}}$.Since the charge is $-Q$,the field points towards the ring. Let the field be $E_1 = \frac{k Q x}{(R^2 + x^2)^{3/2}}$.The electric field due to a point charge $q$ placed at the center at distance $x$ is $E_2 = \frac{k q}{x^2}$.For the net electric field to be zero at $x = R$,we must have $E_1 = E_2$.Substituting $x = R$ into the expression for $E_1$: $E_1 = \frac{k Q R}{(R^2 + R^2)^{3/2}} = \frac{k Q R}{(2R^2)^{3/2}} = \frac{k Q R}{2\sqrt{2} R^3} = \frac{k Q}{2\sqrt{2} R^2}$.Equating $E_1$ and $E_2$: $\frac{k Q}{2\sqrt{2} R^2} = \frac{k q}{R^2}$.Solving for $q$,we get $q = \frac{Q}{2\sqrt{2}}$.

$A$ ring of radius $R$ has charge $-Q$ distributed uniformly over it. The charge that should be placed at the center of the ring such that the electric field becomes zero at a point on the axis of the ring at a distance $R$ from the center of the ring will be:

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