A charge of +3μC is placed in a uniform electric field of strength 1500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
Practice Questions
1 question
Q1
A charge of +3μC is placed in a uniform electric field of strength 1500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
90 J
60 J
30 J
45 J
Work done W = F * d = (E * q) * d = (1500 N/C * 3 × 10^-6 C) * 0.2 m = 0.0009 J = 90 J.
Questions & Step-by-step Solutions
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Q
Q: A charge of +3μC is placed in a uniform electric field of strength 1500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
Solution: Work done W = F * d = (E * q) * d = (1500 N/C * 3 × 10^-6 C) * 0.2 m = 0.0009 J = 90 J.
Steps: 8
Step 1: Identify the given values. We have a charge (q) of +3μC, which is equal to 3 × 10^-6 C, and an electric field strength (E) of 1500 N/C.
Step 2: Convert the charge from microcoulombs to coulombs. 3μC = 3 × 10^-6 C.
Step 3: Identify the distance (d) the charge is moved, which is 0.2 m.
Step 4: Calculate the force (F) acting on the charge using the formula F = E * q. Substitute the values: F = 1500 N/C * 3 × 10^-6 C.
Step 5: Perform the multiplication: F = 1500 * 3 × 10^-6 = 0.0045 N.
Step 6: Now, calculate the work done (W) using the formula W = F * d. Substitute the values: W = 0.0045 N * 0.2 m.
Step 7: Perform the multiplication: W = 0.0045 * 0.2 = 0.0009 J.
Step 8: Convert the work done to a more understandable unit if needed. 0.0009 J is equal to 0.9 mJ (milliJoules).
