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Question

(D) The electric field on the axis of a charged ring at a distance $z_0$ from the center is given by $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q z

Vedclass · Accepted Answer

Both $(a)$ and $(c)$.

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(D) The electric field on the axis of a charged ring at a distance $z_0$ from the center is given by $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q z_0}{(R^2 + z_0^2)^{3/2}}$.Since the particle $P$ has a negative charge $-q$,the force acting on it is $F = -qE = -\frac{1}{4\pi \varepsilon_0} \cdot \frac{Q q z_0}{(R^2 + z_0^2)^{3/2}}$.The negative sign indicates that the force is always directed towards the origin $O$. As the particle crosses the origin,the force reverses direction,always pulling it back towards the center. Thus,the motion is periodic for all $z_0 > 0$.If $z_0 \ll R$,we can approximate $(R^2 + z_0^2)^{3/2} \approx (R^2)^{3/2} = R^3$.Then the force becomes $F \approx -\left( \frac{Q q}{4\pi \varepsilon_0 R^3} \right) z_0$.Since $F \propto -z_0$,the motion is approximately simple harmonic for small $z_0$.

$A$ positively charged thin metal ring of radius $R$ is fixed in the $xy$-plane with its centre at the origin $O$. $A$ negatively charged particle $P$ is released from rest at the point $(0, 0, z_0)$,where $z_0 > 0$. Then the motion of $P$ is:

Similar Questions

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