Electrostatics: Exam-Pattern Drill (2)

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Question

Three point charges of +2+2 μC, +2+2 μC, and 4-4 μC are placed at the vertices of an equilateral triangle of side 0.10.1 m. Find the magnitude of the net electrostatic force on the 4-4 μC charge. Take k=9×109k = 9 \times 10^9 N·m2^2/C2^2.

Solution — Step by Step

The two +2+2 μC charges each attract the 4-4 μC. Each pair force:

F=kq1q2r2=9×109×2×106×4×106(0.1)2F = \frac{k q_1 q_2}{r^2} = \frac{9 \times 10^9 \times 2 \times 10^{-6} \times 4 \times 10^{-6}}{(0.1)^2} F=9×109×8×10120.01=7.2 NF = \frac{9 \times 10^9 \times 8 \times 10^{-12}}{0.01} = 7.2 \text{ N}

So each +2+2 μC pulls the 4-4 μC with 7.27.2 N along the line joining them.

The two +2+2 μC charges sit at vertices of an equilateral triangle, with the 4-4 μC at the third vertex. The two attractive force vectors on the 4-4 μC point from its position toward each of the other two vertices. The angle between these vectors equals the interior angle of the triangle at the 4-4 μC vertex, which is 60°60°.

Fnet=F2+F2+2F2cos60°=F2+2×0.5=F3F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2 \cos 60°} = F\sqrt{2 + 2 \times 0.5} = F\sqrt{3} Fnet=7.2×312.47 NF_{\text{net}} = 7.2 \times \sqrt{3} \approx 12.47 \text{ N}

Final Answer: Fnet12.47F_{\text{net}} \approx 12.47 N (directed along the perpendicular bisector toward the +2+2 μC pair).

Why This Works

Coulomb’s law gives the magnitude of force between two charges; for multiple charges, we use vector superposition. In an equilateral arrangement, the symmetry makes the angle between any two pair-forces equal to 60°60°, which simplifies the parallelogram-law calculation to F3F\sqrt{3}.

The direction of the net force is along the symmetry axis (the perpendicular bisector of the line joining the two like charges), pointing toward them since the test charge is being attracted.

Alternative Method

Resolve each force into x- and y-components along this symmetry axis. The components perpendicular to the axis cancel, and the components along the axis add: Fnet=2Fcos30°=2F(3/2)=F3F_{\text{net}} = 2F \cos 30° = 2F \cdot (\sqrt{3}/2) = F\sqrt{3}. Same answer, different route.

Adding the magnitudes directly (Fnet=7.2+7.2=14.4F_{\text{net}} = 7.2 + 7.2 = 14.4 N) ignores that forces are vectors. This works only if both forces are parallel — never assume that without checking the geometry.

This exact configuration (with sign flips) appears in JEE Main almost every year. The trick to fast solving: spot the symmetry first, then compute one pair-force, then multiply by the right geometric factor (3\sqrt{3} for 60°60°, 2\sqrt{2} for 90°90°).

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