Question

A parallel plate capacitor has plates with dimensions 3 cm by 4 cm separated by 2 mm. The plates are connected across a 60 V battery. Find; i. The capacitance of the capacitor. (3 marks) ii. The charge stored on each plate. (2 marks) iii. The number of electrons transferred in the process. (2 mark) iv. The stored energy in the capacitor.

Answer 1

Answer on Question 82846, Physics, Other

Question:

A parallel plate capacitor has plates with dimensions $3\,\mathrm{cm}$ by $4\,\mathrm{cm}$ separated by $2\,\mathrm{mm}$. The plates are connected across a $60\,\mathrm{V}$ battery. Find:

i) The capacitance of the capacitor

ii) The charge stored on each plate

iii) The number of electrons transferred in the process

iv) The stored energy in the capacitor

Solution:

i) We can find the capacitance of the capacitor from the formula:

$$C = \varepsilon_0 \frac{A}{d'}$$

here, $\varepsilon_0$ is the permittivity of free space, $A$ is the area of overlap of the conducting surfaces, $d$ is the plate separation.

Then, we get:

$$C = \varepsilon_0 \frac{A}{d} = 8.854 \cdot 10^{-12} \frac{F}{m} \cdot \frac{3 \cdot 10^{-2} \, m \cdot 4 \cdot 10^{-2} \, m}{2 \cdot 10^{-3} \, m} = 5.31 \cdot 10^{-12} \, F.$$

ii) We can find the charge stored on each plate from the formula:

$$Q = C \Delta V,$$

here, $C$ is the capacitance of the capacitor, $\Delta V$ is the voltage across the plates of the capacitor.

Then, we get:

$$Q = C \Delta V = 5.31 \cdot 10^{-12} \, F \cdot 60 \, V = 3.2 \cdot 10^{-10} \, C.$$

iii) We can find the number of electrons transferred in the process from the formula:

$$Q = N e,$$

here, $Q$ is the charge stored on each plate of the capacitor, $N$ is the number of electrons transferred in the process, $e = 1.6 \cdot 10^{-19} \, C$ is the charge of the electron.

Then, we get:

$$N = \frac{Q}{e} = \frac{3.2 \cdot 10^{-10} \, C}{1.6 \cdot 10^{-19} \, C} = 2 \cdot 10^9 \, \text{electrons}.$$

iv) We can find the stored energy in the capacitor from the formula:

$$E = \frac{1}{2} C(\Delta V)^2 = \frac{1}{2} \cdot 5.31 \cdot 10^{-12} \, F \cdot (60 \, V)^2 = 9.55 \cdot 10^{-9} \, J.$$

Answer:

i) $C = 5.31 \cdot 10^{-12} \, F.$

ii) $Q = 3.2 \cdot 10^{-10} \, C.$

iii) $N = 2 \cdot 10^9 \, \text{electrons}$.

iv) $E = 9.55 \cdot 10^{-9} \, J.$

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