CEE Kerala Engineering
CEE Kerala Engineering Solved Paper-2010
question_answer
Two point charges\[+5\mu C\]and\[-2\mu C\]are kept at a distance of 1 m in free space. The distance between the two zero potential points on the line joining the charges is
A) \[\frac{2}{7}m\]
B) \[\frac{2}{3}m\]
C) \[\frac{22}{21}m\]
D) \[\frac{20}{21}m\]
E) \[\frac{8}{21}m\]
Correct Answer:
D
Solution :
Let the potential be zero at P and Q. Then solving for\[{{x}_{1}}\] \[\frac{k\times 5}{{{x}_{1}}}=\frac{2k}{(1-{{x}_{1}})}\] \[{{x}_{1}}=\frac{5}{7}\] Similarly, \[\frac{k\times 5}{1+{{x}_{2}}}=\frac{k\times 2}{{{x}_{2}}}\] \[\therefore \] \[{{x}_{2}}=\frac{2}{3}\] Separation\[PQ=\left( 1-\frac{5}{7} \right)+\frac{2}{3}\] \[=\frac{20}{21}\]