Problem

Two positive point charges of $15 \times 10^{-10} \, \mathrm{C}$ and $13 \times 10^{-10} \, \mathrm{C}$ are placed $12 \, \mathrm{cm}$ apart. Find the work done in bringing the two charges $4 \, \mathrm{cm}$ closer.

Data

• $q_1 = 15 \times 10^{-10} \, \mathrm{C}$
• $q_2 = 13 \times 10^{-10} \, \mathrm{C}$
• Initial separation: $r = 12 \, \mathrm{cm} = 0.12 \, \mathrm{m}$
• Change in separation: $\Delta r = 4 \, \mathrm{cm} = 0.04 \, \mathrm{m}$
• Final separation: $R = r - \Delta r = 0.12 \, \mathrm{m} - 0.04 \, \mathrm{m} = 0.08 \, \mathrm{m}$
• Coulomb’s constant: $k = 9 \times 10^9 \, \mathrm{Nm^2C^{-2}}$

Prerequisite Concepts

• Work Done in Moving Charges:

$$W = k q_1 q_2 \left( \frac{1}{r} - \frac{1}{R} \right)$$

Answer

Step 1: Apply the Work Formula

$$W = k q_1 q_2 \left( \frac{1}{r} - \frac{1}{R} \right)$$

Step 2: Substitute Values

$$W = (9 \times 10^9)(15 \times 10^{-10})(13 \times 10^{-10}) \left( \frac{1}{0.12} - \frac{1}{0.08} \right)$$

Step 3: Simplify

$$W = (9)(15)(13) \times 10^{-11} \left( \frac{1}{0.12} - \frac{1}{0.08} \right)$$

Step 4: Evaluate Fractions

$$\frac{1}{0.12} = 8.33, \quad \frac{1}{0.08} = 12.5$$ $$\frac{1}{0.12} - \frac{1}{0.08} = 8.33 - 12.5 = -4.17$$

Step 5: Final Calculation

$$W = (9)(15)(13) \times 10^{-11} \times (-4.17)$$ $$W = -7.318 \times 10^{-8} \, \mathrm{J}$$

Final Answer

The work done in bringing the charges closer is $-7.318 \times 10^{-8} \, \mathrm{J}$.

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