Electrostatics: Numerical Problems Set (7)

Question

Two point charges $q_1 = +4 \mu\text{C}$ and $q_2 = -2 \mu\text{C}$ are placed $30 \text{ cm}$ apart in vacuum. Find (a) the force between them, (b) the electric field at the midpoint of the line joining them, (c) the location of the point on the line where the net field is zero. Take $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.

Solution — Step by Step

Final answers: $F = \mathbf{0.8 \text{ N}}$ attractive, $E_{mid} = \mathbf{2.4 \times 10^6 \text{ N/C}}$, zero point at $\mathbf{\approx 72.4 \text{ cm}}$ from $q_2$ on the far side.

Why This Works

The midpoint trick: when charges have opposite signs, their fields add at every point between them (both push/pull the test charge in the same direction). When charges have the same sign, fields subtract between them — and the zero-field point lies between them.

For zero net field with opposite signs, the point must be outside, on the side of the smaller-magnitude charge. (You’re farther from the bigger charge, so $1/r^2$ scaling can balance the bigger $|q|$.)

Alternative Method

For part (c), use $|q_1|/d_1^2 = |q_2|/d_2^2$ with $d_1/d_2 = \sqrt{|q_1|/|q_2|} = \sqrt{2}$. With $d_1 = d_2 + 0.3$, solve directly: $d_2 = 0.3/(\sqrt{2}-1) \approx 0.724 \text{ m}$.

Common Mistake