Question

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?

Short answer

1. The dielectric constant K of the dielectric is 1.80.
2. Thepotential difference between the plates before and after the dielectric insertion is2.00 V.
3. The electric field at a point midway between the plates before and after the dielectric insertion respectively is $1000N/C$.

Step-by-step solution

(a) Determination of the dielectric constant K of the dielectric.

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.

(b) Determination of the potential difference between the plates before and after the dielectric insertion.

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.

(c) Determination of the dielectric field at a point midway between the plates before and after the dielectric insertion.

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 $1000N/C$.