A charge of +5μC is placed at the origin. What is the electric field at a point

Practice Questions

Q1
A charge of +5μC is placed at the origin. What is the electric field at a point 2m away along the x-axis?
  1. 0 N/C
  2. 1125 N/C
  3. 2250 N/C
  4. 4500 N/C

Questions & Step-by-Step Solutions

A charge of +5μC is placed at the origin. What is the electric field at a point 2m away along the x-axis?
  • Step 1: Identify the charge (q) and its value. Here, q = +5μC, which is equal to 5 × 10^-6 C.
  • Step 2: Identify the distance (r) from the charge to the point where we want to find the electric field. Here, r = 2m.
  • Step 3: Use the formula for electric field (E) due to a point charge: E = k * |q| / r², where k is the electrostatic constant (approximately 9 × 10^9 N m²/C²).
  • Step 4: Substitute the values into the formula: E = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (2 m)².
  • Step 5: Calculate the denominator: (2 m)² = 4 m².
  • Step 6: Now calculate the electric field: E = (9 × 10^9) * (5 × 10^-6) / 4.
  • Step 7: Perform the multiplication: (9 × 10^9) * (5 × 10^-6) = 45,000 N m²/C.
  • Step 8: Finally, divide by 4: E = 45,000 N m²/C / 4 = 11,250 N/C.
  • Step 9: Round the answer to three significant figures: E = 1125 N/C.
  • Coulomb's Law – The electric field due to a point charge is calculated using Coulomb's Law, which states that the electric field (E) is proportional to the charge (q) and inversely proportional to the square of the distance (r) from the charge.
  • Electric Field Direction – Understanding that the electric field direction is away from positive charges and towards negative charges.
  • Units of Measurement – Familiarity with the units used in the calculation, such as microcoulombs (μC) and newtons per coulomb (N/C).
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