Question

Four parallel-plate capacitors are constructed using square plates, and each has a dielectric inserted between the plates.?

Answer 1

Answer:

Explanation:

Capacitance: It is known as the ratio of change in electric charge to the corresponding change in the electric potential.

Write the expression for capacitance of a capacitor.

$C=\frac{\epsilon _0KA }{d}$

Here, electric permittivity is $\epsilon _0$, area of capacitor is $A$, and the distance between the capacitor plates is $d$.

Write the expression for capacitance of A.

$C_A=\frac{\epsilon _0KA }{d}$

Here, electric permittivity is $\epsilon _0$, area of capacitor is $A$, and the distance between the capacitor plates is $d$.

Substitute  for $l^2$ for $A$ .

 $C_A=\frac{\epsilon _0Kl^2A}{d}$

Write the expression for capacitance of B.

$C_B=\frac{\epsilon _0KA }{d}$

Substitute $(\frac{l}{2})^2$ for $A$,  for $(\frac{d}{2})$, and $4K$ for $K$.

$C_B=\frac{\epsilon _0(\frac{l}{2})^24K }{d}\\\\\frac{2\epsilon _0l^2K}{d}$

Write the expression for capacitance of C.

$C_C=\frac{\epsilon _0KA }{d}$

Substitute $(2l)^2$ for $A$ and $2K$ for $K$.

$C_C=\frac{\epsilon _0(2l)^2(2K )}{d}\\\\\frac{8\epsilon _0Kl^2}{d}$

Write the expression for capacitance of D.

$C_D=\frac{\epsilon _0KA }{d}$

 

Substitute $2d$ for $d$ and $2K$ for $K$.

 $C_D=\frac{\epsilon _0 l^2(2K )}{2d}\\\\\frac{\epsilon _0l^2K}{d}$

The relative permittivity and the area of capacitor $A$ are directly proportional to capacitance, and the distance between the capacitor $A$ is inversely proportional to capacitance. As the area of the capacitor is increased, the overall capacitance is increased, and as the distance between the capacitor increases, capacitance decreases. Thus, capacitance depends on the area of the capacitor, distance between the capacitor, and the relative permittivity.

 $C_A=\frac{\epsilon _0Kl^2A}{d}$

$C_B=\frac{2\epsilon _0l^2K}{d}$

$C_C=\frac{8\epsilon _0Kl^2}{d}$

$C_D=\frac{\epsilon _0l^2K}{d}$

$C_C$>$C_B$>$C_A$>$C_D$