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EAMCET Medical EAMCET Medical Solved Paper-2008

  • question_answer
    An oil drop having a charge was kept between two plates having a potential difference of 400 V is in equilibrium. Now another drop of same oil with same charge but double the radius is introduced between the plates. Then the potential difference necessary to keep the drop in equilibrium is

    A)  200 V                                   

    B)  800 V

    C)   1600 V                                

    D)  3200 V

    Correct Answer: D

    Solution :

                     From \[mg={{F}_{e}}\] \[\left( \frac{4}{3}\pi {{r}^{3}} \right)\rho g=qE=q\frac{V}{d}\] \[\Rightarrow \]               \[V\propto \frac{{{r}^{3}}}{q}\] \[\therefore \]  \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{r_{1}^{3}{{q}_{2}}}{r_{2}^{3}{{q}_{1}}}\]         or            \[\frac{400}{{{V}_{2}}}=\frac{r_{1}^{3}q}{{{(2{{r}_{1}})}^{3}}q}\] or            \[\frac{400}{{{V}_{2}}}=\frac{r_{1}^{3}}{8r_{1}^{3}}\] or            \[{{V}_{2}}\,=3200\,\,volt\]


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