A positively charged ring is in the y-z plane | vedclass

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(A) The electric potential $V$ at a distance $x$ from the center of a charged ring of radius $a$ and charge $q$ is given by $V = \frac{kq}{\sqrt{x^2 +

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Increases continuously

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(A) The electric potential $V$ at a distance $x$ from the center of a charged ring of radius $a$ and charge $q$ is given by $V = \frac{kq}{\sqrt{x^2 + a^2}}$.At the origin $(x=0)$,the potential is $V_{center} = \frac{kq}{a}$.As the test charge $q_0$ moves along the $x$-axis away from the origin,the distance $\sqrt{x^2 + a^2}$ increases,which means the potential $V$ decreases.The potential energy $U$ of the test charge is $U = q_0 V$. Since $V$ decreases as $x$ increases,the potential energy $U$ also decreases.According to the law of conservation of energy,the total mechanical energy remains constant. As the potential energy $U$ decreases,the kinetic energy $K$ must increase to compensate.Therefore,the speed of the test charge increases continuously as it moves away from the origin.

$A$ positively charged ring is in the $y-z$ plane with its centre at the origin. $A$ positive test charge $q_0$,held at the origin,is released along the $x$-axis. Then its speed

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