The electric field everywhere on the surface of a charged sphere of radius 0.260 m has a magnitude of 545 N/C and points radially outward from the center of the sphere.
(a) What is the net charge on the sphere?
Use Gauss' Law.
The total flux is
4*pi*R^3 *E
Gauss' Law tells you how to relate the flux to the total charge.
Flux = Q/epsilon0
epsilon0 = 8.854*10^-12 C^2/(n*m^2)
4*pi*R^2 *E
Yes, Damon is right. Typo exponent error. Thanks for catching it, DC
To find the net charge on the sphere, we need to use Gauss's Law, which relates the electric field to the net charge enclosed by a Gaussian surface.
Gauss's Law states that the electric flux through a closed Gaussian surface is equal to the net charge enclosed divided by the permittivity of free space (ε₀).
In this case, the electric field is given as 545 N/C. Since the electric field points radially outward from the center of the sphere, we can assume a Gaussian surface that encloses the entire sphere.
The area of a sphere is given by A = 4πr², where r is the radius. In this case, the radius is given as 0.260 m.
Now, we need to calculate the electric flux through this Gaussian surface. The electric flux (Φ) is given by Φ = E * A * cos(θ), where E is the electric field, A is the area, and θ is the angle between the electric field and the normal to the surface. In this case, since the electric field is radial, θ = 0 degrees, and cos(θ) = 1.
Plugging in the values, we have Φ = 545 N/C * 4π(0.260 m)² * 1.
To find the net charge enclosed, we rearrange Gauss's Law:
Φ = Q_enclosed / ε₀.
Solving for Q_enclosed, we have Q_enclosed = Φ * ε₀.
The permittivity of free space (ε₀) is a constant equal to 8.85 x 10⁻¹² C²/N·m².
Plugging in the values and solving for Q_enclosed, we have:
Q_enclosed = Φ * ε₀ = (545 N/C * 4π(0.260 m)² * 1) * (8.85 x 10⁻¹² C²/N·m²).
Calculating this expression will give us the net charge enclosed by the Gaussian surface, which is the net charge on the sphere.