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How much work(in joules) is done in moving a charge of 2.5μC a distance of 22cm along an equipotential at 10~V?

User Hitch
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Final answer:

The work done in moving a charge of 2.5 μC a distance of 22 cm along an equipotential at 10 V is 2.5 × 10-5 J.

Step-by-step explanation:

The work done in moving a charge of 2.5 μC a distance of 22 cm along an equipotential at 10 V can be calculated using the formula:

Specifically, when a charge of 2.5μC moves a distance of 22cm along an equipotential at 10V, since there's no potential difference, the work done, W, is given by W = qΔV, which equals 2.5μC times 0 V. This results in zero joules of work being done.

Work = charge × potential difference

Substituting the given values, we get:

Work = (2.5 × 10-6 C) × (10 V) = 2.5 × 10-5 J

Therefore, the work done is 2.5 × 10-5 J.

User Malavika
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