What is the potential difference between two points A and B in an electric field if the work done in moving a charge of 5μC from A to B is 0.02 J?

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Q: What is the potential difference between two points A and B in an electric field if the work done in moving a charge of 5μC from A to B is 0.02 J?

Solution: Potential difference V = W/Q = 0.02 J / (5 × 10^-6 C) = 4000 V = 4 V.

Steps: 6

Step 1: Identify the formula for potential difference (V) which is V = W/Q, where W is the work done and Q is the charge.
Step 2: Note the values given in the question: Work done (W) = 0.02 J and Charge (Q) = 5μC.
Step 3: Convert the charge from microcoulombs (μC) to coulombs (C). 5μC = 5 × 10^-6 C.
Step 4: Substitute the values into the formula: V = 0.02 J / (5 × 10^-6 C).
Step 5: Calculate the potential difference: V = 0.02 / 0.000005 = 4000 V.
Step 6: Write the final answer: The potential difference between points A and B is 4000 V.