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, 0)] along the straight path AB is :                                                                                                                                                                                                

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 :

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