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 \[W=({{V}_{A}}-{{V}_{B}})\,(-Q)=0\]You need to login to perform this action.
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