In the capacitor of capacitance 20 mu F- the distance between plates is 2 mm- If a material of dielectric constant 2 is inserted between the plates- then the capacitance of the system is -20 mu F30 mu F22 times 5 mu F40 mu F

Question

In the capacitor of capacitance 20 mu F- the distance between plates is 2 mm- If a material of dielectric constant  2 is inserted between the plates- then the capacitance of the system is -20 mu F30 mu F22 times 5 mu F40 mu F

Toppr Admin · Accepted Answer

We knowC-x3F5-0AdWhen k is addedC1-k-x3F5-0Ad-x21D2-C1-2-xD7-20-x21D2-C1-40-x3BC-F

In the capacitor of capacitance $20 μ F$ , the distance between plates is $2 m m$ . If a material of dielectric constant $2$ is inserted between the plates, then the capacitance of the system is :

$20 μ F$
$30 μ F$
$22 \times 5 μ F$
$40 μ F$

We know
$C = \frac{ϵ_{0} A}{d}$
When k is added
$C_{1} = \frac{k ϵ_{0} A}{d}$
$\Rightarrow C_{1} = 2 \times 20$
$\Rightarrow C_{1} = 40 μ F$

In the capacitor of capacitance 20 μF, the distance between plates is 2 mm. If a material of dielectric constant 2 is inserted between the plates, then the capacitance of the system is :20μF30μF22×5μF40μF

We knowC=ϵ0AdWhen k is addedC1=kϵ0Ad⇒C1=2×20⇒C1=40μF

In the capacitor of capacitance $20 μ F$ , the distance between plates is $2 m m$ . If a material of dielectric constant $2$ is inserted between the plates, then the capacitance of the system is :

$20 μ F$
$30 μ F$
$22 \times 5 μ F$
$40 μ F$

We know
$C = \frac{ϵ_{0} A}{d}$
When k is added
$C_{1} = \frac{k ϵ_{0} A}{d}$
$\Rightarrow C_{1} = 2 \times 20$
$\Rightarrow C_{1} = 40 μ F$