Answer:
a
[tex]E = 25000 \ V/m[/tex]
b
[tex]\sigma = 2.2125* 10^{-7} \ C/m^2[/tex]
c
[tex]q = 2.2125 *10^{-8} \ C [/tex]
d
[tex]C = 4.425 *10^{-10} \ F[/tex]
e
[tex]U = 5.531 *10^{-7} \ J[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 2mm = 0.002 \ m[/tex]
The voltage is [tex]V = 50\ V[/tex]
The area is [tex]A =0.1m^2[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{V}{d}[/tex]
=> [tex]E = \frac{50}{ 0.002}[/tex]
=> [tex]E = 25000 \ V/m[/tex]
Generally the capacitance mathematically represented is
[tex]C = \frac{\epsilon_o * A }{d}[/tex]
Here [tex]\episilon_o[/tex] is the permitivity of free space with value
[tex]\episilon_o = 8.85 * 10^{-12} C/(V⋅m [/tex]
=> [tex]C = \frac{ 8.85 * 10^{-12} * 0.1 }{0.002}[/tex]
=> [tex]C = 4.425 *10^{-10} \ F[/tex]
Generally the charge is mathematically represented as
[tex]q = C * V[/tex]
[tex]q = 4.425 *10^{-10} *50 [/tex]
=> [tex]q = 2.2125 *10^{-8} \ C [/tex]
Generally the charge density is mathematically represented as
[tex]\sigma = \frac{q}{A}[/tex]
=> [tex]\sigma = \frac{ 2.2125 *10^{-8}}{0.1}[/tex]
=> [tex]\sigma = 2.2125* 10^{-7} \ C/m^2[/tex]
Generally the energy stored in this capacitor is mathematically
[tex]U = \frac{1}{2} * CV^2[/tex]
=> [tex]U = \frac{1}{2} * (4.425 *10^{-10} ) * (50)^2[/tex]
=> [tex]U = 5.531 *10^{-7} \ J[/tex]
