For two opposite charges, the neutral point where the net electric field is zero lies outside the charges on the line joining them, closer to the charge of smaller magnitude.

Let the point $P$ be at a distance $x$ from the smaller charge $(-q)$, away from $(+4q)$:

• Distance from $(-q) = x$
• Distance from $(+4q) = r + x$

At point $P$, the magnitudes of electric fields due to both charges must be equal:

$$\frac{1}{4\pi\varepsilon_0} \frac{q}{x^2} = \frac{1}{4\pi\varepsilon_0} \frac{4q}{(r + x)^2}$$ $$\frac{1}{x^2} = \frac{4}{(r + x)^2} \implies \frac{1}{x} = \frac{2}{r + x}$$ $$r + x = 2x \implies x = r$$