A spherical surface completely surrounds a collection of charges. Find the electric flux through the surface if the collection consists of each of the following.
(a) a single +3.70 10-6 C charge
N·m2/C
(b) a single -2.90 10-6 C charge
N·m2/C
(c) both of the charges in (a) and (b)
N·m2/C
flux=integral E.dA and over a closed surface this is eual to the enclosed charge divided by permitivity.
To find the electric flux through a spherical surface surrounding a collection of charges, we can use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is equal to the product of the electric field strength and the area of the surface.
The electric flux, Φ, through a closed surface is given by the equation Φ = E * A, where E is the electric field strength and A is the area of the surface.
In this case, since the surface is spherical, we need to calculate the area of the spherical surface surrounding the charges.
The area of a spherical surface is given by the equation A = 4πr^2, where r is the radius of the sphere.
Let's calculate the electric flux for each case:
(a) For a single +3.70 10-6 C charge:
- The electric flux, Φ = E * A.
- The electric field strength, E, depends on the charge enclosed by the surface and the radius of the sphere.
- Since there is only one charge, the electric field strength is given by E = k * Q / r^2, where k is the Coulomb's constant (9.0 * 10^9 N·m^2/C^2), Q is the charge (3.70 * 10^-6 C), and r is the radius of the sphere.
- The area of the spherical surface, A, is given by A = 4πr^2.
- Putting it all together, the electric flux through the surface is Φ = (k * Q / r^2) * (4πr^2)
(b) For a single -2.90 10-6 C charge:
- The process is the same as in case (a), but with the charge changed to -2.90 * 10^-6 C.
(c) For both charges in (a) and (b):
- Since the electric field due to the charges is a vector quantity, we need to consider the combined effect of both charges.
- The electric field strength, E, at any point outside the charges is given by the vector sum of the electric fields due to each charge individually.
- So, we need to calculate the electric field strength separately for each charge using the same equation as in cases (a) and (b), and then add them together.
- Once we have the combined electric field strength, we can calculate the electric flux using the same equation as in cases (a) and (b).
By following these steps and plugging in the given values for the charges and the radius of the sphere, you will be able to find the electric flux for each case.