Question

Two$10-\mathrm{cm}$-diameter electrodes $0.50\mathrm{cm}$apart form a parallel plate capacitor. The electrodes are attached by metal wires to

the terminals of a $15\mathrm{V}$battery. After a long time, the capacitor is disconnected from the battery but is not discharged. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes

a. Right after the battery is disconnected?

b. After insulating handles are used to pull the electrodes away from each other until they are $1.0\mathrm{cm}$apart?

c. After the original electrodes (not the modified electrodes of part b) are expanded until they are $20\mathrm{cm}$in diameter?

Short answer

(a). - Charge across the capacitor is $0.2\mathrm{nC}$.

- Electric field inside the capacitor is $3000\mathrm{V}/\mathrm{m}$.

- Potential difference between the electrodes is $15\mathrm{V}$.

(b). - Charge across the capacitor is $0.2\mathrm{nC}$.

- Electric field inside the capacitor is $3000\mathrm{V}/\mathrm{m}$.

- Potential difference between the electrodes is $30\mathrm{V}$.

(c). - Charge across the capacitor is $0.8\mathrm{nC}$.

- Electric field inside the capacitor is $3000\mathrm{V}/\mathrm{m}$.

- Potential difference between the electrodes is $15\mathrm{V}$.

Step-by-step solution

part(a) step 1: Given information

Diameter of the electrodes, $D=10\mathrm{cm}=0.1\mathrm{m}$

Distance of electrode from the parallel plate capacitor, $d=0.5\mathrm{cm}=0.005\mathrm{m}$

Voltage of the battery, $V=15\mathrm{V}$

part(a) step 2: Calculation

The electric field strength of the capacitor is given by the formula

$E=\frac{Q}{{\epsilon }_0\Lambda }$

Since the capacitor is connected to the battery for a long time therefore they have the same potential difference as that of battery.

Potential difference across the capacitor,

$\Delta V=15\mathrm{V}$

The electric filed inside the capacitor is calculated by the formula

$$\begin{gathered} V=Ed \\ 15=E\times 0.005 \\ E=3\mathrm{kV}/\mathrm{m} \end{gathered}$$

Charge on each electrode is calculated as

$E=\frac{Q}{L_{0}A}=\frac{Q}{{ϵ}_0\times \frac{{\omega }^2}{4}}$

Plugging the values in the above equation

$$\begin{gathered} 3000=\frac{Q}{8.85\times {10}^{-12}\times \frac{{\text{π×0.1}}^2}{4}} \\ Q=0.2\mathrm{nC} \end{gathered}$$

part(b) step 1: given information

Diameter of the electrodes, $D=10\mathrm{cm}=0.1\mathrm{m}$

Distance of electrode from the parallel plate capacitor, $d=1\mathrm{cm}=0.01\mathrm{m}$Voltage of the battery, $V=15\mathrm{V}$

part(b) step 2: calculation

The electric field strength of the capacitor is given by the formula

$E=\frac{Q}{{\epsilon }_{0}A}$

Electric field inside a capacitor is constant regardless of the distance between the plates, therefore the electric field inside the capacitor is

$E=3000\mathrm{V}/\mathrm{m}$

The potential difference across capacitor is calculated by the formula

$$\begin{gathered} V=Ed \\ V=3000\times 0.01 \\ V=30\mathrm{V} \end{gathered}$$

Charge on the plates of the capacitor is also does not depend on the distance between the plates, therefore charge on the plates of electrode is

$Q=0.2\mathrm{nC}$

part(c) step 1:  given information

Diameter of the electrodes, $D=20\mathrm{cm}=0.2\mathrm{m}$

Distance of electrode from the parallel plate capacitor, $d=0.5\mathrm{cm}=0.005\mathrm{m}$

Voltage of the battery, $V=15\mathrm{V}$

part(c) step 2: calculation

The electric field strength of the capacitor is given by the formula

$E=\frac{Q}{{\epsilon }_{0}A}$

Since the capacitor is connected to the battery for a long time therefore they have the same potential difference as that of battery.

Potential difference across the capacitor,

$\Delta V=15\mathrm{V}$

The electric filed inside the capacitor is calculated by the formula

$$\begin{gathered} V=Ed \\ 15=E\times 0.005 \\ E=3\mathrm{kV}/\mathrm{m} \end{gathered}$$

Charge on each electrode is calculated as

$E=\frac{Q}{{\epsilon }_{0}A}=\frac{Q}{{\epsilon }_0\times \frac{\pi D^2}{4}}$

Plugging the values in the above equation

$$\begin{gathered} 3000=\frac{Q}{8.85\times {10}^{-12}\times \frac{500.22}{4}} \\ Q=0.8\mathrm{nC} \end{gathered}$$