Answer:
The value is [tex]\rho_s = 4.026 *10^{-6} \ C/m^2[/tex]
Explanation:
From the question we are told that
The radius of the inner conductor is [tex]r_1 = 1 \ mm = 0.001 \ m[/tex]
The radius of the outer conductor is [tex]r_2 = 3 \ mm = 0.003 \ m[/tex]
The potential at the outer conductor is [tex]V = 1.5 kV = 1.5 *10^{3} \ V[/tex]
Generally the capacitance per length of the capacitor like set up of the two conductors is
[tex]C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}[/tex]
Here [tex]\epsilon_o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} C/(V \cdot m)[/tex]
=> [tex]C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}[/tex]
=> [tex]C= 50.6 *10^{-12} \ F/m[/tex]
Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge
Generally the line charge density of the outer conductor is mathematically represented as
[tex]\rho_l = C * V[/tex]
=> [tex]\rho_d = 50.6*10^{-12} * 1.5*10^{3}[/tex]
=> [tex]\rho_d = 7.59*10^{-8} \ C/m[/tex]
Generally the surface charge density is mathematically represented as
[tex]\rho_s = \frac{\rho_l }{2 \pi * r_2 }[/tex] here [tex]2 \pi r = (circumference \ of \ outer \ conductor )[/tex]
=> [tex]\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }[/tex]
=> [tex]\rho_s = 4.026 *10^{-6} \ C/m^2[/tex]
