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:http://en.wikipedia.org/wiki/Gauss%27s_law
E*surface area* epsilion=q
To find the net charge on the sphere, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the electric constant.
The electric flux through a closed surface is given by the formula:
Φ = E * A
Where Φ is the electric flux, E is the electric field, and A is the area of the closed surface.
In this case, the electric field has a constant magnitude of 545 N/C and points radially outward from the center of the sphere. This means that the electric field is perpendicular to the surface of the sphere at every point.
The area of the closed surface is given by the formula:
A = 4πr^2
Where A is the area and r is the radius of the sphere.
Substituting the given values, we have:
A = 4π(0.260 m)^2
A = 4π(0.0676 m^2)
A = 0.853 m^2
Now we can calculate the electric flux:
Φ = 545 N/C * 0.853 m^2
Φ = 464.485 N m^2/C
According to Gauss's law, the electric flux is equal to the charge enclosed divided by the electric constant (ε0). The electric constant is approximately 8.85 x 10^-12 C^2/(N m^2).
Φ = (Qenclosed) / ε0
Solving for Qenclosed, we have:
Qenclosed = Φ * ε0
Qenclosed = 464.485 N m^2/C * 8.85 x 10^-12 C^2/(N m^2)
Qenclosed = 4.11 x 10^-9 C
Therefore, the net charge on the sphere is approximately 4.11 x 10^-9 C.