Question

The capacitance of a parallel plate air capacitor is $10 \, \mu $ F. If now the overlapping area of plates is doubled and separation between the plates is halved, the new capacitance will be :

• A. $60\, \mu F$
• B. $10 \,\mu F$
• C. $20 \,\mu F$
• D. $40\, \mu F$

Answer 1

Answer: (D)

Using the relation $C=\frac{\varepsilon_{0} A}{d}$ (Initially) $C=10 \mu F$
Now, $C=\frac{\varepsilon_{0} A}{d}$
$=\frac{\varepsilon_{0} 2 A}{d / 2}=\frac{4 \varepsilon_{0} A}{d}$
[Here, $\left.A=2 A, d=\frac{d}{2}\right]$
So, $\frac{C}{C}=4$
or $C=4 C=4 \times 10=40\,\mu F$