Question

The capacitance of a parallel plate capacitor is $60 \mu F$. If the distance between the plates is tripled and area doubled then new capacitance will be

• A. $10\, \mu F$
• B. $40\, \mu F$
• C. $80\, \mu F$
• D. $100\, \mu F$

Answer 1

Answer: (B)

$\frac{C_{1}}{C_{0}}=\frac{\varepsilon_{0} A_{1}}{d_{1}} \times \frac{d}{\varepsilon_{0} A} $
$\therefore $ where $A_{1}=2 A$
$d_{1}=3 d$
$\frac{C_{1}}{C_{0}}=\frac{2 A}{3d} \times \frac{d}{A}=\frac{2}{3}$
$C_{1}=60 \times \frac{2}{3}$
$=40 \mu F$