Question

A sphere of radius 5.00 cm c m carries charge 3.00 nC n C . Calculate the electric-field magnitude at a distance 4.00 cm c m from the center of the sphere and at a distance 6.00 cm c m from the center of the sphere if the sphere is a solid insulator with the charge spread uniformly throughout its volume.

Answer 1

Answer:

- 8632.1N/C

- 17483.17N/C

Explanation:

Inside electric field magnitude is different of the outside electric field.

- Inside the sphere we have that the electric field is given by:

$E=\frac{Qr}{4\pi \epsilon_oR^3}$

Where Q is the charge of the sphere, R is its radius and e0 is the dielectric permittivity of vacuum.

By replacing (r=5.00cm=5.00*10^{-3}m, e0=8.85*10^{-12}C^2/Nm^2) we get:

$E(r=4.00cm)=\frac{(3.00*10^{-9}C)(0.04m)}{4\pi (8.85*10^{-12}C^2/Nm^2)(0.05m)^3}=8632.1N/C$

- Outside the sphere we have the formula:

$E=\frac{Q}{4\pi \epsilom_or^2}\\\\E(r=6.00cm)=\frac{3.00*10^{-9}C}{4\pi (8.85*10^{-12}C^2/Nm^2)(0.06)^2}=17483.17N/C$

hope this helps!!