A charge of +5 µC is placed in an electric field of 300 N/C. What is the work done in moving the charge 0.2 m in the direction of the field? (2019)

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Q: A charge of +5 µC is placed in an electric field of 300 N/C. What is the work done in moving the charge 0.2 m in the direction of the field? (2019)

Solution: Work done W = F * d = (E * q) * d = (300 N/C * 5 × 10^-6 C) * 0.2 m = 0.3 J.

Steps: 8

Step 1: Identify the given values. We have a charge (q) of +5 µC (microcoulombs) and an electric field (E) of 300 N/C.
Step 2: Convert the charge from microcoulombs to coulombs. 5 µC = 5 × 10^-6 C.
Step 3: Understand the formula for work done (W) in an electric field. The formula is W = F * d, where F is the force and d is the distance moved.
Step 4: Calculate the force (F) using the formula F = E * q. Substitute the values: F = 300 N/C * 5 × 10^-6 C.
Step 5: Perform the multiplication: F = 300 * 5 × 10^-6 = 0.0015 N.
Step 6: Now, use the force to calculate the work done. Substitute F and d into the work formula: W = F * d = 0.0015 N * 0.2 m.
Step 7: Perform the multiplication: W = 0.0015 * 0.2 = 0.0003 J.
Step 8: Convert the work done to a more readable format if necessary. 0.0003 J can also be expressed as 0.3 mJ (millijoules).

A charge of +5 µC is placed in an electric field of 300 N/C. What is the work done in moving the charge 0.2 m in the direction of the field? (2019)

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