A) \[400%\]
B) \[66.6%\]
C) \[33.3%\]
D) \[200%\]
Correct Answer: B
Solution :
Initial capacitance \[C=\frac{{{\varepsilon }_{0}}A}{d}\] When it is half filled by a dielectric of dielectric constant\[K\], then \[{{C}_{1}}=\frac{K{{\varepsilon }_{0}}A}{d/2}=2K\frac{{{\varepsilon }_{0}}A}{d}\] and \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d/2}=\frac{2{{\varepsilon }_{0}}A}{d}\] \[\therefore \] \[\frac{1}{C'}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}=\frac{d}{2{{\varepsilon }_{0}}A}\left( \frac{1}{K}+1 \right)\] \[=\frac{d}{2{{\varepsilon }_{0}}A}\left( \frac{1}{5}+1 \right)\] \[=\frac{6}{10}\frac{d}{{{\varepsilon }_{0}}A}\] \[\therefore \] \[C'=\frac{5{{\varepsilon }_{0}}A}{3d}\] Hence, increase in capacitance \[=\frac{\frac{5}{3}\frac{{{\varepsilon }_{0}}A}{d}-\frac{{{\varepsilon }_{0}}A}{d}}{\frac{{{\varepsilon }_{0}}A}{d}}\] \[=\frac{5}{3}-1=\frac{2}{3}=66.6%\]You need to login to perform this action.
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