Answer :
Answer:
The value is [tex]V = 4.533 \ V [/tex]
Explanation:
From the question we are told that
The dielectric constant is k = 1.5
The area of each plate is [tex]A = 5 \ cm^2 = 0.0005 m^2[/tex]
The distance between the plates is [tex]d= 1 \ mm = 0.001 \ m[/tex]
The charge on the capacitor is [tex]Q = 6*10^{-11} \ C[/tex]
Generally the electric field in a vacuum is mathematically represented as
[tex]E_o = \frac{V_o}{d}[/tex]
Generally [tex]V_o[/tex] is the voltage of the capacitor which is mathematically represented as
[tex]V_o = \frac{Q}{C_o}[/tex]
Here [tex]C_o [/tex] is the capacitance of the capacitor in a vacuum which is mathematically represented as
[tex]C_o = \frac{\epsilon_o * A}{d} [/tex]
Here [tex]epsilon_o [/tex] is a constant with value [tex]epsilon_o= 8.85*10^{-12} C^2 \cdot N^{-1} \cdot m^{-2}[/tex]
=> [tex]C_o = \frac{8.85*10^{-12} * 0.0005 }{0.001} [/tex]
=> [tex]C_o = 4.425 *10^{-12} \ F[/tex]
So
[tex]V_o = \frac{6*10^{-11} }{ 4.425 *10^{-12}}[/tex]
[tex]V_o = 13.6 \ V[/tex]
So
[tex]E_o = \frac{ 13.6}{0.001}[/tex]
=> [tex]E_o = 13600 \ V/m[/tex]
The electric field when the dielectric slab is inserted is mathematically represented as
[tex]E = \frac{E_o}{k}[/tex]
=> [tex]E = \frac{13600}{1.5}[/tex]
=> [tex]E =9067 \ V/m[/tex]
Generally the electric field between the plates is mathematically evaluated as
[tex]E_{N} = E_o - E[/tex]
=> [tex]E_{N} = 13600- 9067[/tex]
=> [tex]E_{N} = 4533 \ V/m[/tex]
Generally the potential difference between the plates is
[tex]V = 4533 * 0.001 [/tex]
=> [tex]V = 4.533 \ V [/tex]