CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    Two conducting spheres of radii 3 cm and 1 cm are separated by a distance of 10 cm in free space. If the spheres are charged to same potential of 10 V each, the force of repulsion between them is

    A)  \[\left( \frac{1}{3} \right)\times {{10}^{-9}}N\] 

    B)         \[\left( \frac{2}{9} \right)\times {{10}^{-9}}N\] 

    C)         \[\left( \frac{1}{9} \right)\times {{10}^{-9}}N\] 

    D)         \[\left( \frac{4}{3} \right)\times {{10}^{-9}}N\]

    E)  \[\left( \frac{2}{3} \right)\times {{10}^{-9}}N\]

    Correct Answer: A

    Solution :

    \[\frac{K{{q}_{1}}}{3}=10,\frac{K{{q}_{2}}}{1}=10\] \[K{{q}_{1}}=30\times {{10}^{-2}}\] \[K{{q}_{2}}=10\times {{10}^{-2}}\] \[F=\frac{K{{q}_{1}}{{q}_{2}}}{{{10}^{-2}}}\] \[=\frac{(30\times {{10}^{-2}})(10\times {{10}^{-2}})}{K\times {{10}^{-2}}}\] \[F=\frac{1}{3}\times {{10}^{-9}}N\]


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