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The capacitance of a variable capacitor joined with a battery of 100 V is changed from 2 μF to 10 μF. What is the change in the energy stored in it?

Asked in GSEB Board July 2017 · Energy stored and energy density

• (1) 2×10⁻² J
• (2) 2.5×10⁻² J
• (3) 6.5×10⁻² J
• (4) 4×10⁻² J

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives 2×10⁻² J. Sanity check: with the battery connected, U = 1/2CV² rises from 1×10⁻² J to 5×10⁻² J, a change of 4×10⁻² J.
• (2) No single slip gives 2.5×10⁻² J. Sanity check: the energy goes from 1×10⁻² J to 5×10⁻² J, so the change is 4×10⁻² J.
• (3) No single slip gives 6.5×10⁻² J; it is more than even the sum of the two energies, 1×10⁻² + 5×10⁻² = 6×10⁻² J. The change is their difference, 4×10⁻² J.

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