What is the potential difference between two points in an electric field if the work done in moving a charge of +2μC between them is 0.04 J?

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Q: What is the potential difference between two points in an electric field if the work done in moving a charge of +2μC between them is 0.04 J?

Solution: V = W/q = 0.04 J / (2 × 10^-6 C) = 20 V.

Steps: 6

Step 1: Identify the formula for potential difference (V) in an electric field, 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: the work done (W) is 0.04 J and the charge (q) is +2μC.
Step 3: Convert the charge from microcoulombs (μC) to coulombs (C). Since 1μC = 10^-6 C, +2μC = 2 × 10^-6 C.
Step 4: Substitute the values into the formula: V = 0.04 J / (2 × 10^-6 C).
Step 5: Perform the division: 0.04 J divided by 2 × 10^-6 C equals 20 V.
Step 6: Conclude that the potential difference between the two points is 20 V.