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

Question

The capacitance of a parallel plate capacitor is 60 μ F. If the distance between the plates is tripled and area doubled then new capacitance will be (A) 10 μ F (B) 40 μ F (C) 80 μ F (D) 100 μ F

Tardigrade Admin · Accepted Answer

(C1/C0)=(ε0 A1/d1) × (d/ε0 A) ∴ where A1=2 A d1=3 d (C1/C0)=(2 A/3d) × (d/A)=(2/3) C1=60 × (2/3) =40 μ F

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

$10 μ F$

$40 μ F$

$80 μ F$

$100 μ F$

$\frac{C _{1}}{C _{0}} = \frac{ε _{0} A _{1}}{d _{1}} \times \frac{d}{ε _{0} A}$
$∴$ where $A_{1} = 2 A$
$d_{1} = 3 d$
$\frac{C _{1}}{C _{0}} = \frac{2 A}{3 d} \times \frac{d}{A} = \frac{2}{3}$
$C_{1} = 60 \times \frac{2}{3}$
$= 40 μ F$

Q. The capacitance of a parallel plate capacitor is 60μF. If the distance between the plates is tripled and area doubled then new capacitance will be

10μF

40μF

80μF

100μF