A) 40\[\mu F\]
B) 30\[\mu F\]
C) 10\[\mu F\]
D) \[\frac{20}{3}\mu F\]
Correct Answer: D
Solution :
Initial capacitance \[C=10\mu F\] \[\frac{{{\varepsilon }_{0}}A}{d}=10\mu F\] when it is filled with two different media then it will like a combination of two capacitors of capacitances \[{{C}_{1}}\] and \[{{C}_{2}}\] in series. \[{{C}_{1}}=\frac{{{K}_{1}}{{\varepsilon }_{0}}A/2}{d}\] \[=\frac{{{K}_{1}}}{2}\left( \frac{{{\varepsilon }_{0}}A}{d} \right)=\frac{2}{2}\times 10=10\mu F\] \[{{C}_{2}}=\frac{{{K}_{2}}}{2}\left( \frac{{{\varepsilon }_{0}}A}{d} \right)=\frac{4}{2}\times 10=20\mu F\] New capacity \[C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\] \[=\frac{10\times 20}{10+20}\] \[=\frac{200}{30}=\frac{20}{3}\mu F\]
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