Q. Consider a parallel plate capacitor of $10\mu F\ (micro - farad)$ with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to
A) $25\ \mu F$
B) $20\ \mu F$
C) $40\ \mu F$
D) $5\ \mu F$
Explanation
(a) $C_{1} = \frac{\varepsilon_{0}\left( \frac{A}{4} \right)}{d},C_{2} = \frac{K\varepsilon_{0}\left( \frac{A}{2} \right)}{d},C_{3} = \frac{\varepsilon_{0}\left( \frac{A}{4} \right)}{d}$
$C_{eq} = C_{1} + C_{2} + C_{3} = \left( \frac{K + 1}{2} \right)\frac{\varepsilon_{0}A}{d} = \left( \frac{4 + 1}{2} \right) \times 10 = 25\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.