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A charge of +10μC is placed in a uniform electric field of strength 500 N/C. Wha

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Question: A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 2m in the direction of the field?

Options:

  1. 10 J
  2. 1 J
  3. 100 J
  4. 0.5 J

Correct Answer: 10 J

Solution: 

Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 2m = 0.01 J.

A charge of +10μC is placed in a uniform electric field of strength 500 N/C. Wha

Practice Questions

Q1
A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 2m in the direction of the field?
  1. 10 J
  2. 1 J
  3. 100 J
  4. 0.5 J

Questions & Step-by-Step Solutions

A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 2m in the direction of the field?
  • Step 1: Identify the charge value, which is +10μC (microcoulombs). Convert this to coulombs: 10μC = 10 × 10^-6 C.
  • Step 2: Identify the strength of the electric field, which is 500 N/C.
  • Step 3: Identify the distance the charge is moved, which is 2 meters.
  • Step 4: Use the formula for work done (W) in an electric field: W = F * d, where F is the force and d is the distance.
  • Step 5: Calculate the force (F) using the formula F = E * q, where E is the electric field strength and q is the charge. So, F = 500 N/C * 10 × 10^-6 C.
  • Step 6: Calculate the force: F = 500 * 10 × 10^-6 = 0.005 N.
  • Step 7: Now, substitute the force (F) and distance (d) into the work done formula: W = F * d = 0.005 N * 2 m.
  • Step 8: Calculate the work done: W = 0.005 * 2 = 0.01 J.
  • Electric Field and Work – Understanding how work is calculated in an electric field using the formula W = F * d, where F is the force on the charge and d is the distance moved.
  • Charge and Electric Field Interaction – Recognizing the relationship between charge, electric field strength, and the force exerted on the charge.
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