Question

A parallel-plate capacitor with no dielectric has a capacitance of $0.5\mu F$. The space between the plates is filled with equal amounts of two dielectric materials of dielectric constants 2 and 3 as shown in the figure. Find the capacitance of the system.

• A. $1.2\mu F$
• B. $1.8\mu F$
• C. $1.25\mu F$
• D. None of these

Answer 1

The correct option is A $1.2\mu F$

From given we can redraw circuit as
Capacitance for upper part
$C_1=\frac{K{\epsilon }_{0}A}{d/2}=\frac{2{\epsilon }_{0}A\times 2}{d}=4\times 0.5\mu F=2\mu F$

Capacitance for lower part
$C_2=\frac{K{\epsilon }_{0}A}{d/2}=\frac{3{\epsilon }_{0}A\times 2}{d}=6\times 0.5\mu F=3\mu F$

Now,
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{l}{C_2}$

$\begin{array}{cc}{C}_{eq}& =\frac{C_1\times C_2}{C_1+C_2}=\frac{2\times 3}{2+3}\\ & =\frac{6}{5}\mu F=1.24\mu F\end{array}$