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