A parallel-plate capacitor has capacitance C= 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 pC. (a) What is the dielectric constant Kof the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

Capacitance is mathematically expressed as,

$C=\frac{Q}{V}$

The capacitor here is connected to the battery and so there is a constant potential difference between the plates.

Whenever the dielectric is inserted, there is a capacitance change by a factor of K.

Therefore,

$$\begin{gathered} \frac{C_{after}}{C_{before}}=\frac{Q_{after}}{Q_{before}} \\ =\frac{45.0pC}{25.0pC} \\ =1.80 \\ =k \end{gathered}$$

Thus, the dielectric constant is 1.80.

The expression for capacitance is,

$C=\frac{{\epsilon }_{0}A}{d}$

Area,

$$\begin{gathered} A=\pi R^2 \\ =\pi {\left(0.0300m\right)}^2 \\ =2.827\times {10}^{-3}{m}^2 \end{gathered}$$

And distance between the plates,

$$\begin{gathered} d=\frac{{\epsilon }_{0}A}{C} \\ =\frac{\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)\left(2.827\times {10}^{-3}{m}^2\right)}{12.5\times {10}^{-12}F} \\ =2.00\times {10}^{-3}m \end{gathered}$$

Before the dielectric insertion,

$$\begin{gathered} V=\frac{Qd}{{\epsilon }_{0}A} \\ =\frac{\left(25.0\times {10}^{-12}\mathrm{C}\right)\left(2.00\times {10}^{-3}\mathrm{m}\right)}{\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)\left(2.827\times {10}^{-3}{\mathrm{m}}^2\right)} \\ =2.00V \end{gathered}$$

After the dielectric insertion, the voltage remains unchanged because the battery remains connected. Thus, the voltage is 2.00 V.

Just as the voltage is unchanged when the battery remains connected, so the electric field is also unchanged before or after the dielectric insertion.

$$\begin{gathered} E=\frac{V}{d} \\ =\frac{2.00V}{2.00\times {10}^{-3}m} \\ =1000N/C \end{gathered}$$

Thus, the electric field is $1000N/C$.