Question

A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 cm from the center of the sphere. If the net flux through the surface of the sphere is 800 N.m2 /C, find q

Answer 1

`q_1 =7.08*10^(-9)C.`

Step-by-step explanation:

Gauss's Law says that the electric flux
`\Phi_E` through a closed surface is directly proportional to the charge
`Q_(enc)` inside it. More precisely,

`$\Phi_E=\oint_S E\cdot dA = (Q_(enc))/(\epsilon_0). $`

This means what is outside this closed surface
`S` does not contribute to the flux through it because field lines that go in must come out, resulting a zero flux from an external charge.

In our context, this means the charge
`q_2` which is outside the sphere will have zero flux through the surface; therefore, Gauss's law will only be concerned with charge
`q_1` which is inside the sphere; Hence,

`$\Phi_E=\oint_S E\cdot dA = (q_1)/(\epsilon_0) = 800 N\cdot m^2/C. $`

Solving for
`q_1` gives

`$ q_1= (800 N\cdot m^2/C)\epsilon_0, $`

`$ q_1= (800 N\cdot m^2/C)*(8.85*10^(-12)C^2/N\cdot m^2) $`

`\boxed{q_1 =7.08*10^(-9)C. }`

which is the charge inside the sphere.