Q. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_{1} = 3\ $and thickness $\frac{d}{3}\ $while the other one has dielectric constant$\ K_{2} = 6\ $and thickness$\frac{2d}{3}.$capacitance of the capacitor is now$\ \ $
A) 1.8 pF
B) 45 pF
C) 40.5 pF
D) 20.25 pF
Explanation
(c) $\ \ \ \ \ \ \ \ \ C = \frac{C_{1}C_{2}}{C_{1} + C_{2}}$ …(i)
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where $\ C_{1} = \frac{K_{1}\varepsilon_{0}A}{d/3}$ …(ii)
And $\ C_{2} = \ \frac{K_{2}\varepsilon_{0}A}{2d/3}$ …(iii)
It is given that $\ \frac{\varepsilon_{0}A}{d} = 9$pF
On substituting Eqs. (ii)and (iii)in Eq. (i), we get the result
$\ \ \ \ \ \ \ \ \ \ \ \ \ C_{eq} = 40.5\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.