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Charges 5 μC and 10 μC are placed 1 m apart. Work done to bring these charges to a distance of 0.5 m from each other is _____. (K = 9×10⁹ N m² C⁻²)

Asked in GUJCET 2011 · Energy of a system of charges

• (1) 9×10⁴ J
• (2) 18×10⁴ J
• (3) 45×10⁻² J
• (4) 9×10⁻¹ J

Step-by-step solution

Why the other options are wrong

• (1) Uses only one charge: (Kq)/r = (9×10⁹×10×10⁻⁶)/1 = 9×10⁴ is the potential (in volts) of the 10 μC charge at 1 m, not a work; it still has to be multiplied by the other charge.
• (2) The same slip at 0.5 m: (9×10⁹×10×10⁻⁶)/(0.5) = 18×10⁴ is a potential in volts, not the work done.
• (4) Takes the final energy only: (Kq₁q₂)/(0.5) = 0.9 J. The work is the change in energy, so the initial 0.45 J must be subtracted.

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