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Difficulty: HardElectrostatics and Electric Charges

A point charge q1=+9.0×10−9 Cq_1 = +9.0 \times 10^{-9}\text{ C} is fixed at the origin (x=0 mx = 0\text{ m}), and a second point charge q2=−4.0×10−9 Cq_2 = -4.0 \times 10^{-9}\text{ C} is fixed on the x-axis at x=0.5 mx = 0.5\text{ m}. At what position xx (in meters) along the x-axis is the net electric field intensity equal to zero?

Answer: 1.5 m

Answer

The net electric field intensity is zero at x=1.5 mx = 1.5\text{ m}.
The correct position is x=1.5 mx = 1.5\text{ m}. At this point, the electric field from +q1+q_1 points in the +x+x direction with magnitude E1=9.0×109×9.0×10−91.52=36 N C−1E_1 = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-9}}{1.5^2} = 36\text{ N C}^{-1}, and the electric field from −q2-q_2 points in the −x-x direction with magnitude E2=9.0×109×4.0×10−91.02=36 N C−1E_2 = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-9}}{1.0^2} = 36\text{ N C}^{-1}. The two vectors are equal in magnitude and opposite in direction, yielding a net electric field of zero.

Step-by-Step Solution

1
Determine the physical region where electric fields can cancel
The point of zero field lies to the right of q2q_2, i.e., x>0.5 mx > 0.5\text{ m}.
Between the charges, the fields due to +q1+q_1 and −q2-q_2 point in the same direction (+x). To the left of q1q_1, q1q_1 is both larger in magnitude and closer, so E1>E2E_1 > E_2 everywhere. Hence, balance can only occur to the right of the smaller magnitude charge q2q_2.
2
Set up the condition for equal electric field magnitudes
\frac{k |q_1|}{x^2} = \frac{k |q_2|}{(x - 0.5)^2}
For the net field to be zero, the vector sum of E1E_1 and E2E_2 must equal zero, meaning their magnitudes must be equal.
3
Substitute values and simplify the algebraic equation
\frac{9.0 \times 10^{-9}}{x^2} = \frac{4.0 \times 10^{-9}}{(x - 0.5)^2} \implies \frac{9}{x^2} = \frac{4}{(x - 0.5)^2}
Coulomb's constant kk and the power factor 10−910^{-9} cancel from both sides.
4
Take square root on both sides to solve for x
\frac{3}{x} = \frac{2}{x - 0.5} \implies 3(x - 0.5) = 2x \implies x = 1.5\text{ m}
Taking the principal square root reduces the quadratic relation to a simple linear equation.

Key Concept

Electric Field Superposition and Zero Field Condition for Point Charges
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Topics

Electrostatics and Electric Charges

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