Increase in Potential Energy:

Initial Potential Energy ($ U_i $):

The initial distance between the charges is $ d = 10 \, \text{cm} = 0.1 \, \text{m} $.

Initial potential energy $ U_i = \frac{K q_1 q_2}{d} $

Final Potential Energy ($ U_f $):

The charge $ q_2 $ is moved towards $ q_1 $ by $ x = 2 \, \text{cm} = 0.02 \, \text{m} $.

The new separation becomes $ d - x = 0.1 \, \text{m} - 0.02 \, \text{m} = 0.08 \, \text{m} $.

Final potential energy $ U_f = \frac{K q_1 q_2}{d - x} $

Increase in Potential Energy ($ \Delta U $):

The increase in potential energy is given by the difference:

$ \Delta U = U_f - U_i = K q_1 q_2 \left( \frac{1}{d - x} - \frac{1}{d} \right) $

$ \Delta U = \frac{K q_1 q_2 x}{d(d - x)} $

$ K = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 $

$ q_1 = 6 \, \mu\text{C} = 6 \times 10^{-6} \, \text{C} $

$ q_2 = 4 \, \mu\text{C} = 4 \times 10^{-6} \, \text{C} $

$ x = 0.02 \, \text{m} $

$ d = 0.1 \, \text{m} $

Performing the calculation:

$ \Delta U = \frac{(9 \times 10^9) \times (6 \times 10^{-6}) \times (4 \times 10^{-6}) \times 0.02}{0.1 \times 0.08} $

Simplifying:

$ \Delta U = \frac{4.32 \times 10^{-3}}{0.008} = 0.54 \, \text{J} $

Therefore, the increase in potential energy of the system is $ 0.54 \, \text{J} $.

Electrostatics

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