In a capacitor of capacitance 20 μ F the distance between the plates is 2 mm. If a dielectric slab of width 2 mm and dielectric constant 2 is inserted between the plates, then the new capacitance will be:

Question

In a capacitor of capacitance 20 μ F the distance between the plates is 2 mm. If a dielectric slab of width 2 mm and dielectric constant 2 is inserted between the plates, then the new capacitance will be: (A) 22 μ F (B) 26.6 μ F (C) 52.2 μ F (D) 40 μ F

Tardigrade Admin · Accepted Answer

Full space between plates is filled with dielectric ⇒ C prime=2 C ⇒ C prime=40 μ F

Q. In a capacitor of capacitance $20 μ F$ the distance between the plates is $2 mm$ . If a dielectric slab of width $2 mm$ and dielectric constant $2$ is inserted between the plates, then the new capacitance will be:

$22 μ F$

$26.6 μ F$

$52.2 μ F$

$40 μ F$

Full space between plates is filled with dielectric
$\Rightarrow C^{'} = 2 C$
$\Rightarrow C^{'} = 40 μ F$

Q. In a capacitor of capacitance 20μF the distance between the plates is 2mm. If a dielectric slab of width 2mm and dielectric constant 2 is inserted between the plates, then the new capacitance will be:

22μF

26.6μF

52.2μF

40μF