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\[PQ=\left( 1-\frac{5}{7} \right)+\frac{2}{3}\]                                 \[=\frac{20}{21}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner