Question

Two point charges q1 and q2 are separated by a distance of 1.5 m. Find the neutral point if q1 = 5 nC and q2 = -2 nC.

Answer 1

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

$$\dfrac{kq_1}{x^2}=-\dfrac{kq_2}{(r-x)^2},$$ $$q_1(r-x)^2=-q_2x^2,$$ $$\dfrac{q_1}{-q_2}=\dfrac{x^2}{(r-x)^2}$$ $$\sqrt{\dfrac{q_1}{-q_2}}=\dfrac{x}{r-x},$$ $$x=\dfrac{r\sqrt{\dfrac{q_1}{-q_2}}}{1+\sqrt{\dfrac{q_1}{-q_2}}},$$ $$x=\dfrac{1.5\ m\cdot\sqrt{\dfrac{5\cdot10^{-9}\ C}{-(-2\cdot10^{-9}\ C)}}}{1+\sqrt{\dfrac{5\cdot10^{-9}\ C}{-(-2\cdot10^{-9}\ C)}}}=0.92\ m.$$