
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
150.3k+ 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.
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