Respuesta :
Answer:
a) The electric potential between the cylinders, [tex]V(L) = \frac{20b - 100a + 80L}{b-a}[/tex]
b) The electric field between the cylinders, [tex]E = \frac{80}{b-a}[/tex]
Explanation:
Voltage at the inner conductor, V₁ = 20 volts
Voltage at the outer conductor, V₂ = 100 volts
Change in potential, [tex]\triangle V = V_{2} - V_{1}[/tex]
[tex]\triangle V = 100 - 20\\\triangle V = 80 volts[/tex]
The change in potential is given by the formula:
[tex]\triangle V = E(b-a)\\80 = E(b-a)[/tex]
The electric field, [tex]E = \frac{80}{b-a}[/tex]
To get the electric potential at a certain point, L from the center:
The distance from the outer conductor to the center = b - L
The electric potential at L will be given by the equation:
[tex]V(L) = V_{b} - V_{center} \\V(L) = 100 - E(b-L)\\E = \frac{80}{b-a} \\V(L) = 100 - \frac{80}{b-a}(b-L)[/tex]
[tex]V(L) = \frac{100(b-a) - 80(b-L)}{b-a} \\V(L) = \frac{100b - 100a -80b + 80L}{b-a} \\V(L) = \frac{20b - 100a + 80L}{b-a}[/tex]
