Answer :
Answer:
(a) σ = 3.41*10⁻7C/m^2
(b) E = 38,530.1 N/C
Explanation:
(a) In order to calculate the resulting surface charge density, you use the following formula:
[tex]\sigma=\frac{Q}{S}[/tex] (1)
σ: surface charge density
Q: charge of the satellite = 3.1 µC = 3.1*10^-6C
S: surface area of the satellite
The satellite has a spherical form, then, the area of the surface is given by:
[tex]S=4\pi r^2[/tex] (2)
r: radius of the satellite = d/2 = 1.7m/2 = 0.85m
You replace the equation (2) into the equation (1) and solve for the surface charge density:
[tex]\sigma=\frac{3.1*10^{-6}C}{4\pi (0.85m)^2}=3.41*10^{-7}\frac{C}{m^2}[/tex]
The surface charge density acquired by the satellite on one orbit is 3.41*10⁻7C/m^2
(b) The electric field just outside the surface is calculate d by using the following formula:
[tex]E=k\frac{Q}{R^2}[/tex] (3)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the satellite = 0.85m
[tex]E=(8.98*10^9Nm^2/C^2)\frac{3.1*10^{-6}C}{(0.85m)^2}=38530.1\frac{N}{C}[/tex]
The magnitude of the electric field just outside the sphere is 38,530.1 N/C