Example 6.2: Dielectric filled capacitor

Question: A parallel plate capacitor has a plate area of $50\,{\rm cm}^2$ and a plate separation of $1.0$ cm. A potential difference of $V_0= 200$ V is applied across the plates with no dielectric present. The battery is then disconnected, and a piece of Bakelite ($K=4.8$) is inserted which fills the region between the plates. What is the capacitance, the charge on the plates, and the potential difference between the plates, before and after the dielectric is inserted?
 
Answer: Before the dielectric is inserted, the space between the plates is presumably filled with air. Since the dielectric constant of air is virtually indistinguishable from that of a vacuum, let us use the vacuum formula (108) to calculate the initial capacitance $C_0$. Thus,

$$C_0=\frac{\epsilon_0\,A}{d} = \frac{(8.85\times 10^{-12})\,(50\times 10^{-4})} {(1\times 10^{-2})} = 4.4\,{\rm pF}.$$

After the dielectric is inserted, the capacitance increases by a factor $K$, which in this case is $4.8$, so the new capacitance $C$ is given by

$$C= K\,C_0 = (4.8)\, (4.4 \times 10^{-12}) = 21\,{\rm pF}.$$

Before the dielectric is inserted, the charge $Q_0$ on the plates is simply

$$Q_0 = C_0 V_0 = (4.4\times 10^{-12})\,(200) = 8.8\times 10^{-10} \,{\rm C}.$$

After the dielectric is inserted, the charge $Q$ is exactly the same, since the capacitor is disconnected, and so the charge cannot leave the plates. Hence,

$$Q=Q_0 = 8.8\times 10^{-10} \,{\rm C}.$$

The potential difference before the dielectric is inserted is given as $V_0= 200$ V. The potential difference $V$ after the dielectric is inserted is simply

$$V = \frac{Q}{C} =\frac{(8.8\times 10^{-10})}{(21\times 10^{-12})} = 42\, {\rm V}.$$

Note, of course, that $V=V_0/K$.