Answer to Question #91878 in Electric Circuits for Priskilla

Question

Two charges +q and +4q are separated a distance of 2r. Use Gauss’ law to find the electric
field at a distance r around the charge +question and 3r around the charge +q

Answer 1

1) According to Gauss's law, the flux provided by the first and the second charge is

$$\Phi_1=\frac{Q_1}{\epsilon_0}=\frac{q}{\epsilon_0},$$ $$\Phi_2=\frac{Q_2}{\epsilon_0}=\frac{4q}{\epsilon_0},$$

and at a distance r between the charges the flux is

$$\Phi=\Phi_1+\Phi_2=\frac{5q}{\epsilon_0}.$$

On the other hand, Gauss's law can be expressed in terms of electric field:

$$E_1=\frac{\Phi_1}{4\pi r^2}=\frac{q}{4\pi\epsilon_0 r^2},$$ $$E_2=\frac{\Phi_2}{4\pi r^2}=\frac{4q}{4\pi\epsilon_0 r^2}.$$

Since the two charges are positive, in the middle between the charges the total field (directed from the larger charge) is

$$E=E_2-E_1=\frac{3q}{4\pi\epsilon_0 r^2}.$$

2) Reasoning the same way, the flux around the charges will be as in the previous problem.

$$\Phi_1=\frac{q}{\epsilon_0},$$ $$\Phi_2=\frac{4q}{\epsilon_0},$$

and for distance of 3r from the smaller charge in direction of the largest charge the field will be

$$E_1=\frac{\Phi_1}{4\pi\cdot9r^2}=\frac{q}{4\pi\epsilon_0\cdot9r^2},$$

and since the charges are 2r from each other, that distance will be r behind the larger charge:

$$E_2=\frac{\Phi_2}{4\pi\cdot r^2}=\frac{4q}{4\pi\epsilon_0\cdot r^2},$$

the total electric field

$$E=E_1+E_2=\frac{37q}{4\pi\epsilon_0\cdot 9r^2}.$$