| A parallel plate capacitor with air between the plates has a capacitance of \[9pF\]. The separation between the plates is \['d'\]. The space between the plates is now filled with two dielectrics. One of the dielectric has dielectric constant \[{{k}_{1}}=3\] and thickness \[d/3\] while the other one has dielectric constant \[{{k}_{2}}=6\] and thickness \[2d/3\] Capacitance of the capacitor is now |
A) \[45\text{ }pF\]
B) \[40.5pF\]
C) \[20.25pF\]
D) \[1.8pF\]
Correct Answer: B
Solution :
| given, \[\frac{{{\in }_{0}}A}{d}=9pF\] |
| \[{{C}_{1}}=\frac{{{\in }_{0}}{{k}_{1}}A}{d/3}=(3\times 3)\frac{{{\in }_{0}}A}{d}=9\times 9=81pF\] |
| and \[{{C}_{2}}=\frac{{{\in }_{0}}{{k}_{2}}A}{2d/3}=\frac{3\times 6}{2}\frac{{{\in }_{0}}A}{d}=9\times 9=81\operatorname{pF}\] Now \[C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{81}{2}=40.5\operatorname{pF}\] |
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