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Difficulty: HardElectrostatics and Electric Charges

Two point charges, q1=+1.6×10−8 Cq_1 = +1.6 \times 10^{-8}\text{ C} and q2=+6.4×10−8 Cq_2 = +6.4 \times 10^{-8}\text{ C}, are fixed in a vacuum at a distance of 0.60 m0.60\text{ m} apart. At what distance from q1q_1 along the line joining the two charges is the net electric field intensity equal to zero?

  1. 0.20 m0.20\text{ m}Answer
  2. B
    0.40 m0.40\text{ m}
  3. C
    0.12 m0.12\text{ m}
  4. D
    0.30 m0.30\text{ m}

Answer

The distance from q1q_1 where the net electric field intensity is zero is 0.20 m0.20\text{ m}.
The net electric field is zero where the magnitudes of the electric fields produced by both charges are equal (E1=E2E_1 = E_2). Setting up kq1x2=kq2(0.60−x)2\frac{k q_1}{x^2} = \frac{k q_2}{(0.60 - x)^2} with q2=4q1q_2 = 4 q_1 yields 1x2=4(0.60−x)2\frac{1}{x^2} = \frac{4}{(0.60 - x)^2}. Taking the square root gives 1x=20.60−x\frac{1}{x} = \frac{2}{0.60 - x}, which yields x=0.20 mx = 0.20\text{ m} from q1q_1.

Step-by-Step Solution

1
Set up the condition for zero net electric field intensity.
The electric field magnitudes produced by q1q_1 and q2q_2 at distance xx from q1q_1 must be equal in magnitude and opposite in direction: E1=E2E_1 = E_2.
Since both charges are positive, the point of zero net electric field must lie on the line segment connecting them.
2
Substitute the electric field formula into the equilibrium equation.
kq1x2=kq2(d−x)2\frac{k q_1}{x^2} = \frac{k q_2}{(d - x)^2}, where d=0.60 md = 0.60\text{ m}.
Electric field intensity due to a point charge is given by E=kqr2E = \frac{k q}{r^2}.
3
Simplify the equation by canceling common terms and substituting known charge values.
1.6×10−8x2=6.4×10−8(0.60−x)2  ⟹  1x2=4(0.60−x)2\frac{1.6 \times 10^{-8}}{x^2} = \frac{6.4 \times 10^{-8}}{(0.60 - x)^2} \implies \frac{1}{x^2} = \frac{4}{(0.60 - x)^2}.
Dividing both sides by k×1.6×10−8k \times 1.6 \times 10^{-8} reduces the numerical coefficients to simple integers.
4
Take the square root of both sides and solve for xx.
1x=20.60−x  ⟹  0.60−x=2x  ⟹  3x=0.60  ⟹  x=0.20 m\frac{1}{x} = \frac{2}{0.60 - x} \implies 0.60 - x = 2x \implies 3x = 0.60 \implies x = 0.20\text{ m}.
Taking the square root removes the quadratic terms and gives a linear relation for the distance xx from q1q_1.

Key Concept

Electric field superposition and point of zero field intensity between like point charges
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Topics

Electrostatics and Electric Charges

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