Question

Two charges +a and +4q are separated by a distance 2r. Use Gauss' law to find the electric field at a distance. R around charge +q

Answer 1

Due to +q charge flux

$\phi_1=\frac{q}{\epsilon_0}$

Due to +4q charge flux

$\phi_1=\frac{4q}{\epsilon_0}$

Net flux

$\phi=\phi_1+\phi_2$

$\phi=\frac{q}{\epsilon_0}+\frac{4q}{\epsilon_0}=\frac{5q}{\epsilon_0}$

Electric field

$E_1=\frac{kQ}{R^2}\\E_2=\frac{4kq}{R^2}$

$E=E_2-E_1$

$E=\frac{4kq}{R^2}-\frac{kq}{R^2}=\frac{3kq}{R^2}$

Net direction (E.F)of electric field E₂ direction