Question

In the figure, a capacitor is filled with dielectrics. The resultant capacitance is

• A. $\frac{2 \varepsilon_{0} A}{d}\left[\frac{1}{k_{1}}+\frac{1}{k_{2}}+\frac{1}{k_{3}}\right]$
• B. $\frac{\varepsilon_{0} A}{d}\left[\frac{1}{k_{1}}+\frac{1}{k_{2}}+\frac{1}{k_{3}}\right]$
• C. $\frac{2 \varepsilon_{0} A}{d}\left[k_{1}+k_{2}+k_{3}\right]$
• D. None of these

Answer 1

Answer: (D)

$C_{1}=\frac{k_{1} \varepsilon_{0} \frac{A}{2}}{\left(\frac{d}{2}\right)}=\frac{k_{1} \varepsilon_{0} A}{d}$
$C_{2}=\frac{k_{2} \varepsilon_{0}\left(\frac{A}{2}\right)}{\left(\frac{d}{2}\right)}=\frac{k_{2} \varepsilon_{0} A}{d}$
and $C_{3}=\frac{k_{3} \varepsilon_{0} A}{2 d}=\frac{k_{3} \varepsilon_{0} A}{2 d}$
Now, $C_{e q}=C_{3}+\frac{C_{1} C_{2}}{C_{1}+C_{2}}$
$=\left(\frac{k_{3}}{2}+\frac{k_{1} k_{2}}{k_{1}+k_{2}}\right) \cdot \frac{\varepsilon_{0} A}{d}$