question_answer

As per this diagram a point charge +q is placed at the origin O. Work done in taking another point charge–Q  from the point A [co-ordinates \[(0,\,\,a)\]] to another points [co-ordinates \[(a,\text{ }0)\]] along the straight path AB is:                  [AIPMT (S) 2005]

A) zero

B) \[\left( \frac{-qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right)\sqrt{2}a\]

C) \[\left( \frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right)\cdot \frac{a}{\sqrt{2}}\]

D) \[\left( \frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right)\sqrt{2}a\]

Correct Answer: A

Solution :

[a] Key Idea: The work done in carrying a test charge consists in product of difference of potentials at points A and B and value of test charge.

Potential at A

\[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{a}\]

Potential at B

\[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{a}\]

Thus, work done in carrying a test charge–Q from A to B.

\[W=({{V}_{A}}-{{V}_{B}})\,(-Q)=0\]