Question

A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?

Answer 1

For the given capacitor,

$$\begin{gathered} C=5\mathrm{\mu F} \\ V=6\mathrm{V} \\ d=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m} \end{gathered}$$

(a) The charge on the positive plate is calculated using q = CV.

Thus,

$q=5\mathrm{\mu F}\times 6\mathrm{V}=30\mathrm{\mu C}$

(b) The electric field between the plates of the capacitor is given by
$E=\frac{V}{d}=3\times {10}^3\mathrm{V}/\mathrm{m}$

(c) Separation between the plates of the capacitor, d = 2×10^$-$3 m
Dielectric constant of the dielectric inserted, k = 5
Thickness of the dielectric inserted, t = 1×10^$-$3 m
Now,
Area of the plates of the capacitor
$$\begin{gathered} \mathrm{C}=\frac{{\in }_0\mathrm{A}}{d} \\ \Rightarrow 5\times {10}^{-6}=\frac{8.85\times {10}^{-12}\times A}{2\times {10}^{-3}} \\ \Rightarrow {10}^4=8.85\times A \\ \Rightarrow A={10}^4\times \frac{1}{8.85} \end{gathered}$$

When the dielectric placed in it, the capacitance becomes
$$\begin{gathered} C_1=\frac{{\in }_{0}A}{d-t+{\displaystyle \frac{t}{k}}} \\ \Rightarrow C_1=\frac{8.85\times {10}^{-12}\times {\displaystyle \frac{{10}^4}{8.85}}}{2\times {10}^{-3}-{10}^{-3}+{\displaystyle \frac{{10}^{-3}}{5}}} \\ \Rightarrow C_1=\frac{8.85\times {10}^{-12}\times {\displaystyle \frac{{10}^4}{8.85}}}{{10}^{-3}+{\displaystyle \frac{{10}^{-3}}{5}}} \\ \Rightarrow C_1=\frac{{10}^{-12}\times {10}^4\times 5}{6\times {10}^{-3}}=8.33\mathrm{\mu F} \end{gathered}$$

(d)
The charge stored in the capacitor initially is given by
$$\begin{gathered} C=5\times {10}^{-6}\mathrm{F} \\ \mathrm{Also}, \\ V=6\mathrm{V} \\ \mathrm{Now}, \\ Q=CV=3\times {10}^{-5}\mathrm{F} \\ \Rightarrow Q=30\mathrm{\mu C} \end{gathered}$$

The charge on the capacitor after inserting the dielectric slab is given by
$$\begin{gathered} C_1=8.3\times {10}^{-6}\mathrm{F} \\ \therefore Q\text{'}=C_{1}V=8.3\times 6\times {10}^{-6} \\ \Rightarrow Q\text{'}=50\mathrm{\mu C} \\ \mathrm{Now}, \\ \mathrm{Charge}\mathrm{flown},Q\text{'}-Q=20\mathrm{\mu C} \end{gathered}$$