A positively charged thin metal ring of radius R is fixed in xy

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1. Electric Charges and Fields

hard

A positively charged thin metal ring of radius $R$ is fixed in the $xy - $ plane with its centre at the $O$. 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

Periodic for all values of ${z_0}$ satisfying $0 < {z_0} < \infty $

Simple harmonic for all values of satisfying $0 < {z_0} < R$

Approximately simple harmonic provided ${z_0} < < R$

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

(IIT-1998)

Solution

(d) Here $E = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{Q{z_0}}}{{{{({R^2} + z_0^2)}^{3/2}}}}$
where $Q$ is the charge on ring and ${z_0}$is the distance of the point from origin.
Then $F = qE = \frac{{ – \,Qq{z_0}}}{{4\pi {\varepsilon _0}{{({R^2} + z_0^2)}^{3/2}}}}$
When charge $-q$ crosses origin, force is again towards centre i.e., motion is periodic.
Now if ${z_0} < < R$
$F = – \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{Qq{z_0}}}{{{R^2}}}$ $⇒$$F \propto – {z_0}$ i.e., motion is $S.H.M.$

Standard 12

Physics

Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of electric field at the point $C$ is

medium

The charge distribution along the semi-circular arc is non-uniform . Charge per unit length $\lambda $ is given as $\lambda = {\lambda _0}\sin \theta $ , with $\theta $ measured as shown in figure. $\lambda_0$ is a positive constant. The radius of arc is $R$ . The electric field at the center $P$ of semi-circular arc is $E_1$ . The value of $\frac{{{\lambda _0}}}{{{ \in _0}{E_1}R}}$ is

hard

A positively charged thin metal ring of radius $R$ is fixed in the $xy - $ plane with its centre at the $O$. 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

Solution

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