What is the amount of work done in moving a charge of 6 C from a point with a potential of 120 V to a point with a potential of 140 V?
120 J
To find the amount of work done in moving a charge, we can use the formula for electric work done: \(W = q \times \Delta V\), where \(W\) is the work done, \(q\) is the charge in coulombs, and \(\Delta V\) is the change in electric potential (voltage) in volts.
Given:
- Charge, \(q = 6 \, \text{C}\)
- Initial potential, \(V_1 = 120 \, \text{V}\)
- Final potential, \(V_2 = 140 \, \text{V}\)
The change in potential, \(\Delta V\), is calculated as:
\(\Delta V = V_2 - V_1 = 140 \, \text{V} - 120 \, \text{V} = 20 \, \text{V}\)
Substituting the known values into the formula:
\(W = 6 \, \text{C} \times 20 \, \text{V} = 120 \, \text{J}\)
Therefore, the amount of work done is 120 J, which corresponds to the correct answer.
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