Question

a) You have two charges 3q and -3q placed at some distance r. Where do you think that the electric field will be zero? WHy do you think so?

b) 3q and 2q charges are placed at some distance. Where you would place third charges q so that the new force on this charge is zero? Why did you place it there?

Answer 1

(a) The electric field will be zero at point $x$ where $E_1=E_2$:

$$\dfrac{kq_1}{x^2}=\dfrac{kq_2}{(r-x)^2},$$ $$q_1(r-x)^2=q_2x^2,$$ $$3q(r-x)^2=-3qx^2,$$ $$2x^2-2xr+r^2=0.$$

This quadratic equation has only imaginary roots. Therefore, there is no zero-field point for a pair of equal magnitude but opposite sign charges.

(b) The electric forces on the charge $q$ due to charges $3q$ and $2q$ must be balanced:

$$F_1=F_2,$$ $$\dfrac{kq_1q}{x^2}=\dfrac{kq_2q}{(r-x)^2},$$ $$\dfrac{k3q^2}{x^2}=\dfrac{k2q^2}{(r-x)^2},$$ $$\dfrac{3}{x^2}=\dfrac{2}{(r-x)^2},$$ $$3(r-x)^2=2x^2,$$ $$\dfrac{3}{2}=\dfrac{x^2}{(r-x)^2},$$ $$\sqrt{\dfrac{3}{2}}=\dfrac{x}{(r-x)},$$ $$x=\dfrac{r\sqrt{\dfrac{3}{2}}}{(1+\sqrt{\dfrac{3}{2}})}.$$