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 [coordinates \[(0,a)\]]

A) to another point B [coordinates \[(a,0)\]] along the straight path AB is zero                      

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

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

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

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