A charge of +5 μC is placed in an electric field of 1000 N/C. What is the work done in moving the charge 0.2 m in the direction of the field? (2022)
Practice Questions
1 question
Q1
A charge of +5 μC is placed in an electric field of 1000 N/C. What is the work done in moving the charge 0.2 m in the direction of the field? (2022)
1 J
0.1 J
0.5 J
2 J
Work done W = F * d = (E * q) * d = (1000 N/C * 5 × 10^-6 C) * 0.2 m = 0.1 J.
Questions & Step-by-step Solutions
1 item
Q
Q: A charge of +5 μC is placed in an electric field of 1000 N/C. What is the work done in moving the charge 0.2 m in the direction of the field? (2022)
Solution: Work done W = F * d = (E * q) * d = (1000 N/C * 5 × 10^-6 C) * 0.2 m = 0.1 J.
Steps: 10
Step 1: Identify the given values. We have a charge (q) of +5 μC (which is 5 × 10^-6 C) and an electric field (E) of 1000 N/C.
Step 2: Convert the charge from microcoulombs to coulombs. +5 μC = 5 × 10^-6 C.
Step 3: Identify the distance (d) the charge is moved, which is 0.2 m.
Step 4: Use the formula for work done (W) in an electric field: W = F * d, where F is the force on the charge.
Step 5: Calculate the force (F) using the formula F = E * q. Substitute the values: F = 1000 N/C * 5 × 10^-6 C.
Step 6: Calculate the force: F = 1000 * 5 × 10^-6 = 0.005 N.
Step 7: Now, substitute the force (F) and distance (d) into the work formula: W = F * d = 0.005 N * 0.2 m.
Step 8: Calculate the work done: W = 0.005 * 0.2 = 0.001 J.
Step 9: Correctly calculate the work done using the correct formula: W = (E * q) * d = (1000 N/C * 5 × 10^-6 C) * 0.2 m.
Step 10: Final calculation: W = (1000 * 5 × 10^-6) * 0.2 = 0.1 J.
