A ring of radius $R$ carries a charge $ + q$ . a test charge $ - {q_o}$ is released on its axis at a distance $\sqrt {3R} $ from its center. How much kinetic energy will be acquired by the test charge when it reaches the center of the ring?
(A) $\dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{q{q_o}}}{R}$
(B) $\dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{q{q_o}}}{{2R}}$
(C) $\dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{q{q_o}}}{{\sqrt 3 R}}$
(D) $\dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{q{q_o}}}{{3R}}$
Answer
300.9k+ views
Hint The electric potential at a point is defined as the amount of work done on a unit charge to bring it from infinity or an unknown point to a reference point. Find the potential at the initial distance, that is, the distance from which the test charge was released. Find the potential at the final point or the potential at the center of the ring. From these two potentials, find the kinetic energy acquired by the test charge.
Complete step by step answer
Let the potential at the initial point be denoted by ${V_i}$ and the potential at the final point, the potential at the center be denoted by ${V_f}$ . The potential ${V_i}$ is given as
${V_i} = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{{\sqrt {{R^2} + {{\left( {\sqrt 3 R} \right)}^2}} }}$
$ \Rightarrow {V_i} = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{{2R}}$
Now, we will find the potential at the center of the ring.
${V_f} = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{R}$
Therefore, we can define kinetic energy as the test charge multiplied by the potential difference between the initial and final points. That is,
$KE = {q_o}({V_f} - {V_i})$
By substituting the calculated values of initial and final potentials into the above equation, we get
$KE = {q_o}\left( {\dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{R} - \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{{2R}}} \right)$
Evaluating the above equation gives us the kinetic energy acquired by the test charge gives us
$KE = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{q{q_o}}}{{2R}}$
Therefore, this is the value of kinetic energy acquired by the test charge.
Hence, we can conclude the option (B) to be the correct option.
Note
Note that for calculating the potential at the initial point, we have not directly used the given value of the distance from the center of the ring as we need the distance from the charges on the ring or the distance from the ring itself. So we have used the Pythagoras theorem to compute the distance from the initial point of the test charge to the ring.
Complete step by step answer
Let the potential at the initial point be denoted by ${V_i}$ and the potential at the final point, the potential at the center be denoted by ${V_f}$ . The potential ${V_i}$ is given as
${V_i} = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{{\sqrt {{R^2} + {{\left( {\sqrt 3 R} \right)}^2}} }}$
$ \Rightarrow {V_i} = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{{2R}}$
Now, we will find the potential at the center of the ring.
${V_f} = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{R}$
Therefore, we can define kinetic energy as the test charge multiplied by the potential difference between the initial and final points. That is,
$KE = {q_o}({V_f} - {V_i})$
By substituting the calculated values of initial and final potentials into the above equation, we get
$KE = {q_o}\left( {\dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{R} - \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{{2R}}} \right)$
Evaluating the above equation gives us the kinetic energy acquired by the test charge gives us
$KE = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{q{q_o}}}{{2R}}$
Therefore, this is the value of kinetic energy acquired by the test charge.
Hence, we can conclude the option (B) to be the correct option.
Note
Note that for calculating the potential at the initial point, we have not directly used the given value of the distance from the center of the ring as we need the distance from the charges on the ring or the distance from the ring itself. So we have used the Pythagoras theorem to compute the distance from the initial point of the test charge to the ring.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

How to Convert a Galvanometer into an Ammeter or Voltmeter

Hybridisation in Chemistry – Concept, Types & Applications

What Are Elastic Collisions in One Dimension?

Effective Nuclear Charge for JEE

