CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric of dielectric constant 4 is inserted between these two charges. If the thickness of the slab is r/2, then the force between the charges will become:

    A)  \[F\]                    

    B)  \[\frac{3}{5}F\]

    C)  \[\frac{4}{9}F\]               

    D)         \[\frac{F}{4}\]

    E)  \[\frac{F}{2}\]

    Correct Answer: D

    Solution :

    From Coulombs law the force (F) between two charges is \[F=\frac{1}{4\pi {{\varepsilon }_{0}}k}\frac{{{q}^{2}}}{{{r}^{2}}}\] First case:    \[k=1\]                 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{{{r}^{2}}}\]             Second case:   \[k=4\]                 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}\times 4}.\frac{{{q}^{2}}}{{{r}^{2}}}\]            ?. (2) \[\Rightarrow \]               \[\frac{F}{F}=4\] \[\Rightarrow \]               \[F=\frac{F}{4}\]


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