Question

A parallel plate capacitor is constructed with a dielectric slab with κ = 1.5 inserted between the plates. The area of each plate is 5 cm2, and the distance between the two plates is 1 mm. Assume the infinite plane approximation. 1) If we fix the charge on the capacitor to be Q = 6×10-11 C, what is the potential difference between the top and bottom plates? Note that in the region containing the dielectric medium, the electric field is E = E0 / κ, where E0 is the electric field in vacuum.

Answer 1

Answer:

The value is $V = 4.533 \ V$

Explanation:

From the question we are told that

The dielectric constant is k = 1.5

The area of each plate is $A = 5 \ cm^2 = 0.0005 m^2$

The distance between the plates is $d= 1 \ mm = 0.001 \ m$

The charge on the capacitor is $Q = 6*10^{-11} \ C$

Generally the electric field in a vacuum is mathematically represented as

$E_o = \frac{V_o}{d}$

Generally $V_o$ is the voltage of the capacitor which is mathematically represented as

$V_o = \frac{Q}{C_o}$

Here $C_o$ is the capacitance of the capacitor in a vacuum which is mathematically represented as

$C_o = \frac{\epsilon_o * A}{d}$

Here $epsilon_o$ is a constant with value $epsilon_o= 8.85*10^{-12} C^2 \cdot N^{-1} \cdot m^{-2}$

=> $C_o = \frac{8.85*10^{-12} * 0.0005 }{0.001}$

=> $C_o = 4.425 *10^{-12} \ F$

So

$V_o = \frac{6*10^{-11} }{ 4.425 *10^{-12}}$

$V_o = 13.6 \ V$

So

$E_o = \frac{ 13.6}{0.001}$

=> $E_o = 13600 \ V/m$

The electric field when the dielectric slab is inserted is mathematically represented as

$E = \frac{E_o}{k}$

=> $E = \frac{13600}{1.5}$

=> $E =9067 \ V/m$

Generally the electric field between the plates is mathematically evaluated as

$E_{N} = E_o - E$

=> $E_{N} = 13600- 9067$

=> $E_{N} = 4533 \ V/m$

Generally the potential difference between the plates is

$V = 4533 * 0.001$

=> $V = 4.533 \ V$