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Difficulty: HardElectrostatics and Electric Charges

Two point charges, Q1=+3.0×10−6 CQ_1 = +3.0 \times 10^{-6}\text{ C} and Q2=+4.0×10−6 CQ_2 = +4.0 \times 10^{-6}\text{ C}, are positioned in a vacuum at coordinates (0 m,3.0 m)(0\text{ m}, 3.0\text{ m}) and (3.0 m,0 m)(3.0\text{ m}, 0\text{ m}) respectively on a Cartesian plane. Taking the electrostatic constant k=9.0×109 N m2 C−2k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}, what is the magnitude of the net electric field intensity at the origin (0,0)(0,0)?

  1. A
    1.0×103 N C−11.0 \times 10^3\text{ N C}^{-1}
  2. 5.0×103 N C−15.0 \times 10^3\text{ N C}^{-1}Answer
  3. C
    7.0×103 N C−17.0 \times 10^3\text{ N C}^{-1}
  4. D
    1.5×104 N C−11.5 \times 10^4\text{ N C}^{-1}

Answer

The magnitude of the net electric field intensity at the origin is 5.0×103 N C−15.0 \times 10^3\text{ N C}^{-1}.
The electric field intensity at the origin is a vector sum of the individual electric fields created by each point charge. The field due to the charge on the y-axis points downward along the y-axis with a magnitude of 3.0×103 N C−13.0 \times 10^3\text{ N C}^{-1}, while the field due to the charge on the x-axis points leftward along the x-axis with a magnitude of 4.0×103 N C−14.0 \times 10^3\text{ N C}^{-1}. Because these two fields act at right angles to each other, their vector sum magnitude is given by (3.0×103)2+(4.0×103)2=5.0×103 N C−1\sqrt{(3.0 \times 10^3)^2 + (4.0 \times 10^3)^2} = 5.0 \times 10^3\text{ N C}^{-1}.

Step-by-Step Solution

1
Calculate the electric field intensity E1E_1 at the origin due to charge Q1Q_1
E1=k∣Q1∣r12=(9.0×109)(3.0×10−6)3.02=3.0×103 N C−1E_1 = \frac{k |Q_1|}{r_1^2} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})}{3.0^2} = 3.0 \times 10^3\text{ N C}^{-1}, directed along the negative y-axis.
Electric field magnitude follows Coulomb's law for field strength, and positive charges create fields directed away from themselves.
2
Calculate the electric field intensity E2E_2 at the origin due to charge Q2Q_2
E2=k∣Q2∣r22=(9.0×109)(4.0×10−6)3.02=4.0×103 N C−1E_2 = \frac{k |Q_2|}{r_2^2} = \frac{(9.0 \times 10^9)(4.0 \times 10^{-6})}{3.0^2} = 4.0 \times 10^3\text{ N C}^{-1}, directed along the negative x-axis.
The charge is located at (3.0,0)(3.0, 0) on the x-axis, producing a field pointing toward the origin.
3
Calculate the net electric field vector magnitude at the origin
Enet=E12+E22=(3.0×103)2+(4.0×103)2=5.0×103 N C−1E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{(3.0 \times 10^3)^2 + (4.0 \times 10^3)^2} = 5.0 \times 10^3\text{ N C}^{-1}.
Since the two component fields are perpendicular along orthogonal axes (x and y), their resultant is found using the Pythagorean theorem.

Key Concept

Vector addition of electric field intensities from multiple point charges
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Topics

Electrostatics and Electric Charges

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