Two point charges — find the point where electric field is zero

Question

Two point charges $q_1 = +4\mu C$ and $q_2 = +1\mu C$ are placed 30 cm apart. Find the point on the line joining them where the net electric field is zero.

(JEE Main 2023, similar pattern)

Solution — Step by Step

Both charges are positive, so their fields point AWAY from each charge. For the fields to cancel, the test point must be between the two charges? No — between them, both fields point in the same direction (away from each charge, towards each other).

The null point must be outside, on the side of the smaller charge. Beyond $q_2$ , the field from $q_1$ (pointing away from $q_1$ , i.e., to the right) and the field from $q_2$ (pointing away from $q_2$ , also to the right) are in opposite directions from the test point’s perspective. Wait — let us think again.

Actually, beyond $q_2$ (to the right of $q_2$ ): field due to $q_1$ points to the right, field due to $q_2$ also points to the right. Same direction. Between them: $q_1$ pushes right, $q_2$ pushes left — opposite directions but $q_1$ is stronger nearby.

The null point is outside, beyond the smaller charge $q_2$ . Here, $q_2$ ‘s field is strong (we are close) and points left, while $q_1$ ‘s field is weaker (we are far) and points right.

Let the null point $P$ be at distance $x$ from $q_2$ on the side away from $q_1$ . Then $P$ is at distance $(30 + x)$ cm from $q_1$ .

\frac{kq_1}{(30 + x)^2} = \frac{kq_2}{x^2}

\frac{4}{(30 + x)^2} = \frac{1}{x^2}

\frac{(30 + x)^2}{x^2} = 4

\frac{30 + x}{x} = 2 \quad (\text{taking positive root})

30 + x = 2x

x = 30 \text{ cm}

The null point is 30 cm from $q_2$ , on the side away from $q_1$ (i.e., 60 cm from $q_1$ ).

Why This Works

At the null point, the electric fields from both charges have equal magnitude but opposite direction, so they cancel perfectly. The null point is always closer to the smaller charge because a weaker charge needs less distance to match the field of a stronger charge at greater distance.

For two charges of the same sign: the null point is outside, beyond the smaller charge. For two charges of opposite sign: there is no null point between them or outside. Actually, for opposite signs, the null point is on the line extended beyond the larger charge — but these problems are less common.

Alternative Method

You can also use the ratio directly. At the null point, $\frac{r_1}{r_2} = \sqrt{\frac{q_1}{q_2}} = \sqrt{\frac{4}{1}} = 2$ . So $r_1 = 2r_2$ . With $r_1 - r_2 = 30$ (since $P$ is outside): $2r_2 - r_2 = 30$ , giving $r_2 = 30$ cm. This ratio method is the fastest approach.

For JEE and NEET MCQs, the ratio method is a 10-second solution. The ratio of distances from the null point to the two charges equals the square root of the ratio of charges. Remember: $r_1/r_2 = \sqrt{q_1/q_2}$ .

Common Mistake

The most frequent error: placing the null point BETWEEN the two like charges. Between two positive charges, both fields point from the charges towards each other — they add up, not cancel. The null point for like charges is always on the outer side of the smaller charge. Draw the field directions before setting up the equation.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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