Problem

If $280 \, \mathrm{J}$ of work is done in carrying a charge of $2 \, \mathrm{C}$ from a place where the potential is $-12 \, \mathrm{V}$ to another place where the potential is $V$, calculate the value of $V$.

Data

• Work: $W = 280 \, \mathrm{J}$
• Charge: $q = 2 \, \mathrm{C}$
• Initial Potential: $V_A = -12 \, \mathrm{V}$

Prerequisite Concepts

• Potential Difference:

$$\Delta V = \frac{W}{q}$$

• Final Potential:

$$V_B = V_A + \Delta V$$

Answer

Step 1: Calculate Potential Difference

$$\Delta V = \frac{W}{q}$$

Substitute values:

$$\Delta V = \frac{280}{2} = 140 \, \mathrm{V}$$

Step 2: Calculate Final Potential

$$V_B = V_A + \Delta V$$ $$V_B = -12 + 140$$ $$V_B = 128 \, \mathrm{V}$$

Final Answer

The final potential is $V_B = 128 \, \mathrm{V}$.

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