Question

A charge +q is placed at the origin of X-Y axes as shown in the figure. The work done in taking a charge Q from A to B along the straight line AB is:

A. $\dfrac { qQ }{ 4\pi { \epsilon }_{ 0 } } \left( \dfrac { a-b }{ ab } \right)$
B. $\dfrac { qQ }{ 4\pi { \epsilon }_{ 0 } } \left( \dfrac { b-a }{ ab } \right)$
C. $\dfrac { qQ }{ 4\pi { \epsilon }_{ 0 } } \left( \dfrac { b }{ { a }^{ 2 } } -\dfrac { 1 }{ b } \right)$
D. $\dfrac { qQ }{ 4\pi { \epsilon }_{ 0 } } \left( \dfrac { a }{ { b }^{ 2 } } -\dfrac { 1 }{ b } \right)$

Answer 1

Hint: This problem can be solved using the expression for electric potential on a charge. Obtain the expression for electric potential at point A and then for point B. Work done in moving a charge can be calculated by subtracting the potential at initial position from the potential at final position and multiplying it by total charge. This will give you the expression for work done in taking a charge Q from A to B.

Formula used:
$V=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { q }{ r }$
$W=Q\left( { V }_{ B }-{ V }_{ A } \right)$

Complete step by step answer:
Electric potential on a charge is given by,
$V=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { q }{ r }$
Thus, Electric potential at point A is
${V}_{A}=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { q }{ a }$ …(1)
Similarly, electric potential at point B is
${V}_{B}=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { q }{ b }$ …(2)
Now, work done in moving charge Q from A to B is given by,
$W=Q\left( { V }_{ B }-{ V }_{ A } \right)$ …(3)
Substituting values from the equation. (1) and equation. (2) in equation. (3) we get,
$W=Q\left( \dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { q }{ b } -\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { q }{ a } \right)$
$\therefore W=\dfrac { qQ }{ 4\pi { \epsilon }_{ 0 } } \left( \dfrac { 1 }{ b } -\dfrac { 1 }{ a } \right)$
$\therefore W=\dfrac { qQ }{ 4\pi { \epsilon }_{ 0 } } \left( \dfrac { a-b }{ ab } \right)$
Thus, the work done in taking the charge Q from A to B is $\dfrac { qQ }{ 4\pi { \epsilon }_{ 0 } } \left( \dfrac { a-b }{ ab } \right).$
Hence, the correct answer is option A i.e. $\dfrac { qQ }{ 4\pi { \epsilon }_{ 0 } } \left( \dfrac { a-b }{ ab } \right).$

Note:
In electric potential, the charge is moved without any acceleration. As the unit charge was potential, we had to work against electrostatic force. Thus, the potential is positive. If the charge was negative, then we had to work along the electrostatic force. In, that case, potential would have been negative. S.I. unit of electric potential is volt. Volt is defined as Joule per Coulomb.