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Difficulty: MediumElectrostatics and Electric Charges

Two point charges, q1=+2.0×10−6 Cq_1 = +2.0 \times 10^{-6}\text{ C} and q2=−2.0×10−6 Cq_2 = -2.0 \times 10^{-6}\text{ C}, are fixed in a vacuum separated by a distance of 0.20 m0.20\text{ m}. What is the magnitude of the net electric field intensity at the midpoint along the line joining the two charges? (Take Coulomb's constant k=9.0×109 N m2 C−2k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2})

  1. A
    0.0 N C−10.0\text{ N C}^{-1}
  2. B
    9.0×105 N C−19.0 \times 10^5\text{ N C}^{-1}
  3. C
    1.8×106 N C−11.8 \times 10^6\text{ N C}^{-1}
  4. 3.6×106 N C−13.6 \times 10^6\text{ N C}^{-1}Answer

Answer

3.6×106 N C−13.6 \times 10^6\text{ N C}^{-1}
The correct answer is 3.6×106 N C−13.6 \times 10^6\text{ N C}^{-1}. At the midpoint (r=0.10 mr = 0.10\text{ m}), the electric field due to the positive charge points towards the negative charge, and the field due to the negative charge also points towards the negative charge. Summing both equal field magnitudes of 1.8×106 N C−11.8 \times 10^6\text{ N C}^{-1} yields 3.6×106 N C−13.6 \times 10^6\text{ N C}^{-1}.

Step-by-Step Solution

1
Determine the distance from each point charge to the midpoint.
The distance r=0.20 m2=0.10 mr = \frac{0.20\text{ m}}{2} = 0.10\text{ m}.
The midpoint divides the total separation distance equally.
2
Calculate the magnitude of the electric field intensity E1E_1 created by the positive charge q1q_1 at the midpoint.
E1=k∣q1∣r2=9.0×109×2.0×10−6(0.10)2=1.8×106 N C−1E_1 = \frac{k |q_1|}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.10)^2} = 1.8 \times 10^6\text{ N C}^{-1} directed away from q1q_1 (towards q2q_2).
Electric field vectors point away from positive charges.
3
Calculate the magnitude of the electric field intensity E2E_2 created by the negative charge q2q_2 at the midpoint.
E2=k∣q2∣r2=9.0×109×2.0×10−6(0.10)2=1.8×106 N C−1E_2 = \frac{k |q_2|}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.10)^2} = 1.8 \times 10^6\text{ N C}^{-1} directed towards q2q_2.
Electric field vectors point towards negative charges.
4
Combine the electric field vectors vectorially to find the net field intensity.
Enet=E1+E2=1.8×106+1.8×106=3.6×106 N C−1E_{\text{net}} = E_1 + E_2 = 1.8 \times 10^6 + 1.8 \times 10^6 = 3.6 \times 10^6\text{ N C}^{-1}.
Because both E1E_1 and E2E_2 point in the exact same direction (towards q2q_2), their magnitudes add directly.

Key Concept

Superposition Principle of Electric Fields
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Topics

Electrostatics and Electric Charges

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