Examkin

Question

Difficulty: Very hardElectrostatics and Electric Charges

Two point charges, q1=+4.0×10−8 Cq_1 = +4.0 \times 10^{-8}\text{ C} and q2=−9.0×10−8 Cq_2 = -9.0 \times 10^{-8}\text{ C}, are fixed in a vacuum at a distance of 0.50 m0.50\text{ m} apart. A third point charge q3=+2.0×10−8 Cq_3 = +2.0 \times 10^{-8}\text{ C} is placed along the line passing through q1q_1 and q2q_2 such that the net electrostatic force acting on it is zero. What is the distance of q3q_3 from q1q_1?

  1. A
    0.20 m0.20\text{ m}
  2. B
    0.40 m0.40\text{ m}
  3. 1.00 m1.00\text{ m}Answer
  4. D
    1.50 m1.50\text{ m}

Answer

The distance of the third charge from q1q_1 is 1.00 m1.00\text{ m} (located on the side of q1q_1 opposite to q2q_2).
The correct distance of 1.00 m1.00\text{ m} is determined by recognizing that zero net force on a test charge q3q_3 occurs outside the opposite charges q1q_1 and q2q_2, specifically on the side of the smaller charge q1q_1. Equating electrostatic forces gives 4d2=9(d+0.50)2\frac{4}{d^2} = \frac{9}{(d+0.50)^2}, which yields d=1.00 md = 1.00\text{ m}.

Step-by-Step Solution

1
Determine the equilibrium region for the third charge.
Because q1=+4.0×10−8 Cq_1 = +4.0 \times 10^{-8}\text{ C} and q2=−9.0×10−8 Cq_2 = -9.0 \times 10^{-8}\text{ C} have opposite charges, electrostatic forces on q3q_3 point in opposite directions only outside the segment connecting them. To balance the forces, q3q_3 must be closer to the smaller magnitude charge q1q_1, placing it at distance dd to the left of q1q_1.
Between two opposite charges, the force from the positive charge and the force from the negative charge act in the same direction, so net zero force is impossible between them.
2
Set up Coulomb's Law equilibrium equation.
k∣q1q3∣d2=k∣q2q3∣(d+0.50)2  ⟹  4.0×10−8d2=9.0×10−8(d+0.50)2\frac{k |q_1 q_3|}{d^2} = \frac{k |q_2 q_3|}{(d + 0.50)^2} \implies \frac{4.0 \times 10^{-8}}{d^2} = \frac{9.0 \times 10^{-8}}{(d + 0.50)^2}
Equilibrium requires the force magnitude exerted by q1q_1 on q3q_3 to equal the force magnitude exerted by q2q_2 on q3q_3.
3
Solve the equation for distance dd.
4d2=9(d+0.50)2  ⟹  2d=3d+0.50  ⟹  2(d+0.50)=3d  ⟹  d=1.00 m\frac{4}{d^2} = \frac{9}{(d + 0.50)^2} \implies \frac{2}{d} = \frac{3}{d + 0.50} \implies 2(d + 0.50) = 3d \implies d = 1.00\text{ m}.
Taking the square root of both sides simplifies the inverse-square relation to a solvable linear relation.

Key Concept

Electrostatic equilibrium and vector force cancellation for point charges
Rate this question

Topics

Electrostatics and Electric Charges

More questions from this topic