A hollow sphere has a uniform volume charge density of 2.22 nC/m3. The inner radius is a = 12.2 cm and the outer radius is b = 24.4 cm.
What is the magnitude of the electric field at 18.3 cm from the center of the sphere?
What is the magnitude of the electric field at 78.1 cm from the center of the sphere?
Shells can be treated as point charges so this question simplifies considerably. E = kQ/r^2.
To find Q multiply the charge density times the volume. Which by the way is a bit silly because all the charge is on the shell. Whatever.
2.22 * 4/3 pi r^3. You can even make it easier just jumping to E = k *2.22 * 4/3* pi *r. Are you sure there wasn't a charge on the outer shell? It would make things harder
45h
To find the magnitude of the electric field at a certain distance from the center of the sphere, you can use Gauss's Law. Gauss's Law states that the electric field through a closed surface is directly proportional to the charge enclosed by that surface.
To apply Gauss's Law, you need to consider a Gaussian surface, which is a hypothetical closed surface that surrounds the charge distribution. In this case, since we have a hollow sphere, the natural choice for the Gaussian surface would be a sphere centered at the center of the original sphere and with a radius equal to the distance from the center where we want to find the electric field.
For the first question, at a distance of 18.3 cm from the center of the sphere, the Gaussian surface has a radius of 18.3 cm. To find the electric field at this distance, we need to calculate the charge enclosed by this Gaussian surface.
Since the sphere has a uniform volume charge density, the charge enclosed by the Gaussian surface can be calculated by multiplying the volume charge density by the volume enclosed by the Gaussian surface. The volume enclosed by the Gaussian surface can be obtained by subtracting the volume of the inner sphere (a smaller hollow sphere with radius 'a') from the volume of the outer sphere (the given hollow sphere with radius 'b').
The volume of a sphere is given by the equation V = (4/3) * π * r^3. So, the volume of the inner sphere is (4/3) * π * a^3 and the volume of the outer sphere is (4/3) * π * b^3. Subtracting the volume of the inner sphere from the volume of the outer sphere gives us the volume of the hollow part of the sphere, which is the volume enclosed by the Gaussian surface.
Once you have the enclosed charge, you can then use Gauss's Law to find the electric field. Gauss's Law states that the electric field E through a closed surface is given by E = Q_enclosed / (ε0 * A), where Q_enclosed is the charge enclosed by the surface, ε0 is the permittivity of free space (8.854 × 10^-12 C^2 / N·m^2), and A is the area of the Gaussian surface.
Substituting the value of the enclosed charge and the area of the Gaussian surface into Gauss's Law will give you the magnitude of the electric field at a distance of 18.3 cm from the center of the sphere.
Similarly, for the second question, you can follow the same procedure to find the magnitude of the electric field at a distance of 78.1 cm from the center of the sphere, using a Gaussian surface with a radius of 78.1 cm.