| Consider a partially filled capacitor. |
 |
| Half of the volume is filled with a dielectric of susceptibility \[{{X}_{E}}=3.\]Percentage of total energy stored in the dielectric is |
A) 100%
B) 60%
C) 20%
D) 50%
Correct Answer: C
Solution :
| \[\frac{{{U}_{D}}}{{{U}_{Total}}}=\frac{{{U}_{D}}}{{{U}_{gap}}+{{U}_{\mathsf{D}}}}\] |
| \[=\frac{{{U}_{D}}Vo{{l}_{D}}}{{{U}_{gap}}Vo{{\operatorname{l}}_{gap}}+{{U}_{D}}Vo{{l}_{D}}}\] |
| \[\frac{\frac{1}{2}{{\in }_{0}}\frac{E_{0}^{2}}{K}\frac{V}{2}}{\left( \frac{V}{2}\times \frac{1}{2}{{\in }_{0}}{{E}^{2}}+\frac{v}{2}\times \frac{1}{2}\frac{{{\in }_{0}}{{E}^{2}}}{K} \right)}\]\[=\frac{\frac{1}{K}}{1+\frac{1}{K}}=\frac{1}{1+K}=\frac{1}{5}\]\[\Rightarrow {{U}_{D}}=20%U[\therefore K=1+X=1+3=4]\] |
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