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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    A soap bubble is charged to a potential of 16 V. Its radius is, then doubled. The potential of the bubble now will be:

    A)  16V                      

    B)         8V

    C)  4V                         

    D)         2V

    E)  zero

    Correct Answer: B

    Solution :

    Potential on bubble, \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] \[\therefore \]  \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{r}_{2}}}{{{r}_{1}}}\] \[\Rightarrow \]               \[\frac{16}{{{V}^{2}}}=\frac{2}{1}\] \[\Rightarrow \]               \[{{V}_{2}}=8V\]


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