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 0.1m in the direction of the field?
Practice Questions
1 question
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 0.1m in the direction of the field?
0.5 J
1 J
2 J
0.1 J
Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 0.1 m = 0.5 J.
Questions & Step-by-step Solutions
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Q
Q: 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 0.1m in the direction of the field?
Solution: Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 0.1 m = 0.5 J.
Steps: 9
Step 1: Identify the given values. We have a charge (q) of +10μC, which is equal to 10 × 10^-6 C, and an electric field strength (E) of 500 N/C.
Step 2: Convert the charge from microcoulombs to coulombs. +10μC = 10 × 10^-6 C.
Step 3: Identify the distance (d) the charge is moved, which is 0.1 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 = 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 * 0.1 m.
Step 8: Calculate the work done: W = 0.005 * 0.1 = 0.0005 J.
Step 9: Convert the work done to Joules: W = 0.5 J.
