A point charge q = +3.0 μC sits at the origin. What is the electric field at a point 0.50 m to the right along the +x axis?

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Study for the UCF PHY2054 General Physics Exam. Use flashcards and multiple-choice questions complete with hints and explanations. Boost your understanding and get exam-ready!

Multiple Choice

A point charge q = +3.0 μC sits at the origin. What is the electric field at a point 0.50 m to the right along the +x axis?

Explanation:
The electric field from a point charge follows the inverse-square law: E = k q / r^2, directed radially away from the charge if the charge is positive. Here, the charge is +3.0 μC at the origin, and we’re 0.50 m to the right along the +x axis, so the field points in the +x direction. Compute the magnitude: q = 3.0 × 10^-6 C, r = 0.50 m, r^2 = 0.25 m^2, and k ≈ 8.99 × 10^9 N·m^2/C^2. E = (8.99 × 10^9)(3.0 × 10^-6) / 0.25 = (2.697 × 10^4) / 0.25 ≈ 1.08 × 10^5 N/C. Therefore, the electric field is about 1.08 × 10^5 N/C in the +x direction.

The electric field from a point charge follows the inverse-square law: E = k q / r^2, directed radially away from the charge if the charge is positive. Here, the charge is +3.0 μC at the origin, and we’re 0.50 m to the right along the +x axis, so the field points in the +x direction.

Compute the magnitude: q = 3.0 × 10^-6 C, r = 0.50 m, r^2 = 0.25 m^2, and k ≈ 8.99 × 10^9 N·m^2/C^2. E = (8.99 × 10^9)(3.0 × 10^-6) / 0.25 = (2.697 × 10^4) / 0.25 ≈ 1.08 × 10^5 N/C.

Therefore, the electric field is about 1.08 × 10^5 N/C in the +x direction.

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