Question

What is the area of the plates of a $3\, F$ parallel plate capacitor, if the separation between the plates is $5\, mm$ ?

• A. $9.281 \times 10^{9} m ^{2}$
• B. $4.529 \times 10^{9} m ^{2}$
• C. $1.694 \times 10^{9} m ^{2}$
• D. $12.981 \times 10^{9} m ^{2}$

Answer 1

Answer: (C)

Let the area be A
The separation between the plates $d = 5\, mm = 0.005\, m$
We know capacitance $C=\frac{\varepsilon_{0}A}{d}$
$\Rightarrow A=\frac{Cd}{\varepsilon_{0}}$
$= 4\pi \times 3 \times 0.005 \times 9 \times 10^{9}$
$=1.69 \times10^{9}\, m^{2}$