Question

A parallel plate capacitor has d = 1 mm and C = 1 F with no medium inside will have :

• A. Area = 36 $\pi$ x $10^6$
• B. Area = 4 $\pi$ × $10^6$
• C. Area = 6 $\pi$× $10^6$
• D. Area = $\pi$ × $10^6$

Answer 1

Answer: (A)

Area = 36 $\pi$ x $10^6$

Answer 2

The capacitance of a parallel plate capacitor with no medium inside (vacuum or air) is given by the formula:

$C = \frac{\epsilon_0 A}{d}$

where $C$ is the capacitance, $\epsilon_0$ is the permittivity of free space, $A$ is the area of each plate, and $d$ is the separation between the plates.

We are given:

$C = 1$ F $d = 1$ mm $= 1 \times 10^{-3}$ m

We need to find the area $A$. Rearranging the formula, we get:

$A = \frac{C d}{\epsilon_0}$

The value of $\epsilon_0$ is approximately $8.854 \times 10^{-12}$ F/m. However, the options are given in terms of $\pi$, which suggests using the value of $\epsilon_0$ related to the Coulomb constant $k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2$.

From this relationship, we can find $\epsilon_0$:

$\epsilon_0 = \frac{1}{4 \pi k} = \frac{1}{4 \pi \times 9 \times 10^9} = \frac{1}{36 \pi \times 10^9} = \frac{10^{-9}}{36 \pi}$ F/m.

Now, substitute the values of $C$, $d$, and $\epsilon_0$ into the formula for $A$:

$A = \frac{(1 \text{ F}) \times (1 \times 10^{-3} \text{ m})}{\frac{10^{-9}}{36 \pi} \text{ F/m}}$ $A = \frac{1 \times 10^{-3}}{\frac{10^{-9}}{36 \pi}} \text{ m}^2$ $A = \frac{10^{-3} \times 36 \pi}{10^{-9}} \text{ m}^2$ $A = 36 \pi \times 10^{-3 - (-9)} \text{ m}^2$ $A = 36 \pi \times 10^6 \text{ m}^2$