A) \[8{{q}_{2}}\]
B) \[8{{q}_{1}}\]
C) \[6{{q}_{2}}\]
D) \[6{{q}_{1}}\]
Correct Answer: A
Solution :
Key Idea: The change in potential energy of the system is \[{{U}_{D}}-{{U}_{C}}\] as discussed under. When charge \[{{q}_{3}}\] is at C, then its potential energy is \[{{U}_{C}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{q}_{1}}\,{{q}_{3}}}{0.4}+\frac{{{q}_{2}}\,{{q}_{3}}}{0.5} \right)\] When charge \[{{q}_{3}}\] is at D, then \[{{U}_{D}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{q}_{1}}\,{{q}_{3}}}{0.4}+\frac{{{q}_{2}}\,{{q}_{3}}}{0.1} \right)\] Hence, change in potential energy \[\Delta U={{U}_{D}}-{{U}_{C}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{q}_{2}}\,{{q}_{3}}}{0.1}-\frac{{{q}_{2}}\,{{q}_{3}}}{0.5} \right)\] but \[\Delta U=\frac{{{q}_{3}}}{4\pi {{\varepsilon }_{0}}}k\] \[\therefore \,\frac{{{q}_{3}}}{4\pi {{\varepsilon }_{0}}}k=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\left( \frac{{{q}_{2}}\,{{q}_{3}}}{0.1}-\frac{{{q}_{2}}\,{{q}_{3}}}{0.5} \right)\] \[\Rightarrow k={{q}_{2}}\,(10-2)=8{{q}_{2}}\]
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