Answer to Question #91847 in Electric Circuits for Shali

Question

Two charges +q and +4q are separated a distance of 2r. Use Gauss’ law to find the electric
field

Answer 1

The total or net flux out of a closed surface (Gauss's law):

$$\phi_{net}=\oint_S E_n dA=\frac{Q_{inside}}{\epsilon_0}$$

The electric field for a surface in the form of a sphere of radius R (S=4πR²):

$$E=\frac{1}{4\pi\epsilon_0}\frac{Q_{inside}}{R^2}$$

thus possible options:

1. inside the surface there is none of the two charges:

$$Q_{inside}=0 \space \to \space E=0$$

2. inside the surface, you can position a sphere of radius R₁ so that the charge +q would be inside this sphere:

$$E=\frac{1}{4\pi\epsilon_0}\frac{q}{R_1^2}$$

if the charge +q is located in the center of the sphere, then R₁ <2r

3. inside the surface, you can position a sphere of radius R₂ so that the charge +4q would be inside this sphere:

$$E=\frac{1}{4\pi\epsilon_0}\frac{4q}{R_2^2}=\frac{1}{\pi\epsilon_0}\frac{q}{R_2^2}$$

if the charge +4q is located in the center of the sphere, then R₂ <2r

4. inside the surface you can position a sphere of radius R₃>r so that both charges would be inside this sphere:

$$E=\frac{1}{4\pi\epsilon_0}\frac{4q+q}{R_3^2}=\frac{5}{4\pi\epsilon_0}\frac{q}{R_3^2}$$