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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    Two charges of magnitude\[4\times {{10}^{-3}}C\]and \[6\times {{10}^{-8}}C\]at A and B are 50 cm apart. The electric potential is zero at a point along AB, where distance from A is given by

    A)  40 cm           

    B)  20 cm

    C)  10 cm            

    D)  30 cm

    Correct Answer: B

    Solution :

     Suppose, potential at point P is zero. \[{{V}_{1}}={{V}_{2}}\] \[K\times \frac{4\times {{10}^{-8}}}{{{r}_{1}}}=K\times \frac{6\times {{10}^{-8}}}{{{r}_{2}}}\] \[\frac{4\times {{10}^{-8}}}{x}=\frac{6\times {{10}^{-8}}}{(50-x)}\] \[4(50-x)=6x\] \[200-4x=6x\] \[\Rightarrow \] \[10x=200\] \[x=20\text{ }cm\]


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