A parallel plate capacitor with air between the plates has a capacitance of 10 pF. The capacitance, if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 4 is

Question

A parallel plate capacitor with air between the plates has a capacitance of 10 pF. The capacitance, if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 4 is (A) 80 pF (B) 96 pF (C) 100 pF (D) 120 pF

Tardigrade Admin · Accepted Answer

Here, C1=(ε0A/d)=10 PF ldots(i) and C2=k(ε0A/d 2)=(4×2ε0A/d) (∵ k=4) =8×10 (using (i)) =80 PF

Q. A parallel plate capacitor with air between the plates has a capacitance of $10 pF$ . The capacitance, if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant $4$ is

$80 pF$

$96 pF$

$100 pF$

$120 pF$

Here, $C_{1} = \frac{ε _{0} A}{d} = 10 PF \dots (i)$
and $C_{2} = k \frac{ε _{0} A}{d 2} = \frac{4 \times 2 ε _{0} A}{d} (∵ k = 4)$
$= 8 \times 10 (u s in g (i))$
$= 80 PF$

Q. A parallel plate capacitor with air between the plates has a capacitance of 10pF. The capacitance, if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 4 is

80pF

96pF

100pF

120pF