A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} d,$ where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:

(Given C₀ = capacitance of capacitor with air as medium between plates.)

Option 1 -

$\frac{4 K C_{0}}{3 + K}$

Option 2 -

$\frac{3 K C_{0}}{3 + K}$

Option 3 -

$\frac{3 + K}{4 K C_{0}}$

Option 4 -

$\frac{K}{4 + K}$

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Correct Option - 1

Detailed Solution:

$C_{0} = \frac{ε_{0} A}{d}$

$C_{1} = \frac{4 ε_{0} A}{d} = 4 C_{0}$

$C_{Z} = \frac{k ε_{0} A}{\frac{3}{4} d} = \frac{4 k c_{0}}{3}$

Series

$C_{a} = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$

$\frac{16 k}{3 d} \frac{C_{0}^{2}}{4 C_{0} (1 + \frac{k}{3})}$

$= \frac{4 k C_{0}}{k + 3}$

0 Upvotes 0 Downvotes

Similar Questions for you

Four metallic plates, each with a surface area A, are placed at a distance d from each other as shown in the figure. If points A and B is connected with a battery of emf V, then the charge on the plate a will be

alok kumar singh

If q₁+ q₂ = 3q, then the value of the q₁/q₂, for which the force of repulsion between q₁ and q₂ is maximum is

alok kumar singh

For force between q₁ and q₂ is maximum, q₁ = q₂

A capacitor of 20 μF is charged to 500 volt and a capacitor 10μF is charged to 200 volt and connected. Then after connecting both of them, the common potential for each will be :–

alok kumar singh

Three charges Q,(+q) and (+q) are placed at the vertices of an equilateral triangle of side ℓ as shown in the figure. If the net electrostatic energy of the system is zero, then Q is equal to :-