Question

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. $5\, \mu F$

Answer 1

Answer: (A)

$C_{1}=\frac{\varepsilon_{0}\left(\frac{A}{4}\right)}{d},$
$C_{2}=\frac{K \varepsilon_{0}\left(\frac{A}{2}\right)}{d}$

$C_{3} =\frac{\varepsilon_{0}\left(\frac{A}{4}\right)}{d}$
$C_{\text {eq }} =C_{1}+C_{2}+C_{3}$
$=\left(\frac{K+1}{2}\right) \frac{\varepsilon_{0} A}{d}$
$=\left(\frac{4+1}{2}\right) \times 10$
$=25\, \mu F$