Electrostatics: Application Problems (1)

Question

Two point charges $q_1 = +4\,\mu\text{C}$ and $q_2 = -1\,\mu\text{C}$ are placed $30\,\text{cm}$ apart. At what point on the line joining them (and on which side) is the electric field zero?

Solution — Step by Step

The null point is at $30\,\text{cm}$ beyond $q_2$, on the side opposite to $q_1$.

Why This Works

Between the two charges, both fields point the same way (from $+q_1$ towards $-q_2$), so they add — no chance of zero. Beyond $q_1$, the closer $+4\,\mu\text{C}$ always dominates the farther $-1\,\mu\text{C}$. Only beyond $q_2$, where the smaller magnitude charge is closer, can the fields cancel.

We took square roots cleanly because $|q_1|/|q_2| = 4$, giving a factor of $2$. When the ratio isn’t a perfect square, the algebra is messier but the logic is identical.

Alternative Method

Plug in coordinates. Place $q_1$ at $x=0$ and $q_2$ at $x = 0.30$. Test the region $x > 0.30$:

$\frac{4}{x^2} = \frac{1}{(x-0.30)^2}$

Solving gives $x = 0.60\,\text{m}$, which is $0.30\,\text{m}$ beyond $q_2$. Same answer.

Common Mistake

Setting $x$ between the charges and getting an algebraic answer that looks reasonable but is physically impossible. Always check the direction of fields in the region you’ve chosen before equating magnitudes.