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

Question

The capacitance of a parallel plate air capacitor is 10 μ F. If now the overlapping area of plates is doubled and separation between the plates is halved, the new capacitance will be : (A) 60 μ F (B) 10 μ F (C) 20 μ F (D) 40 μ F

Tardigrade Admin · Accepted Answer

Using the relation C=(ε0 A/d) (Initially) C=10 μ F Now, C=(ε0 A/d) =(ε0 2 A/d / 2)=(4 ε0 A/d) [Here, .A=2 A, d=(d/2)] So, (C/C)=4 or C=4 C=4 × 10=40 μ F

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

$60 μ F$

$10 μ F$

$20 μ F$

$40 μ F$

Using the relation $C = \frac{ε _{0} A}{d}$ (Initially) $C = 10 μ F$
Now, $C = \frac{ε _{0} A}{d}$
$= \frac{ε _{0} 2 A}{d /2} = \frac{4 ε _{0} A}{d}$
[Here, $A = 2 A, d = \frac{d}{2}]$
So, $\frac{C}{C} = 4$
or $C = 4 C = 4 \times 10 = 40 μ F$

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

60μF

10μF

20μF

40μF