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

Answer:

$q_1 =7.08*10^{-9}C.$

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,

$\Phi_E=\oint_S E\cdot dA = \dfrac{Q_{enc}}{\epsilon_0}.$

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,

$\Phi_E=\oint_S E\cdot dA = \dfrac{q_1}{\epsilon_0} = 800 N\cdot m^2/C.$

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.