91Ó°ÊÓ

The signs of the charge distribution on the plates are shown in the figure below where the plates A and C are connected to the negative point b while the plate B is connected to the positive point a

We want to find the capacitance C between points a and b. The plates A and B are considered to be a capacitor and the plates B and C is another. Both these capacitors are connected in parallel as shown by the figure. So the total capacitance of the system will be given by

$C=C_{AB}+C_{BC}$ (1)

Where the capacitance of two parallel plates with a dielectric material is given by an equation in the form

$C=\frac{K{\mathrm{Î\mu }}_{oA}}{d}$

So equation (1) will be in the form

$$\begin{gathered} C=\frac{K{\mathrm{Î\mu }}_{oA}}{d}+\frac{K{\mathrm{Î\mu }}_{oA}}{d} \\ C=2\frac{K{\mathrm{Î\mu }}_{oA}}{d} \end{gathered}$$ (2)

Now we can plug our values for K, A, and d into equation (2) to get C where A = 12 cm x 12 cm

$$\begin{gathered} \mathrm{C}=\frac{2{\mathrm{°­Î\mu }}_{\mathrm{o}}\mathrm{A}}{\mathrm{d}} \\ =\frac{2\left(4.20\right)\left(8.854×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\left(144×{10}^{-4}{\mathrm{m}}^2\right)}{0.00045\mathrm{m}} \\ =2.4\mathrm{nF} \end{gathered}$$