Two point charges $q_1 = 6 μ\text{C}$ and $q_2 = 4 μ\text{C}$ are kept at points A and B in air where distance $AB = 10$ cm. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$, by 2 cm ? $(\frac{1}{4πε_0} = 9\times 10^9 \text{SI units})$

Question

Two charges $ q_1 = 2 \, \mu \mathrm{C} $ and $ q_2 = -3 \, \mu \mathrm{C} $ are placed 10 cm apart in a vacuum. What is the magnitude and direction of the force between them?

Answer 1

Step 1: Factor out:
$\Delta U = kq_1q_2\left(\dfrac1{r_2} - \dfrac1{r_1}\right)$ with $kq_1q_2 = 0.216$ J m.

Step 2: Brackets:
$\dfrac1{0.08} - \dfrac1{0.10} = 12.5 - 10 = 2.5$ per metre.

Step 3: Product:
$0.216\times2.5 = 0.54$ J, option (C).

Final Answer:
The potential energy rises by 0.54 J. \[ \boxed{\text{(C) }0.54\ \text{J}} \]

Two point charges \(q_1 = 6 μ\text{C}\) and \(q_2 = 4 μ\text{C}\) are kept at points A and B in air where distance \(AB = 10\) cm. What is the increase in potential energy of the system when \(q_2\) is moved towards \(q_1\), by 2 cm ? \((\frac{1}{4πε_0} = 9\times 10^9 \text{SI units})\)

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The Correct Option is C

Solution and Explanation

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