Practice portal › Electric Potential and Capacitance › Capacitor Circuits and Energy

Work done in placing a charge of 8×10⁻¹⁸ C on a capacitor of capacitance 800 μF is

Asked in GUJCET 2007 · Energy stored and energy density

• (1) 4×10⁻³² J
• (2) 32×10⁻³² J
• (3) 3.1×10⁻²⁶ J
• (4) 16×10⁻³² J

Step-by-step solution

Why the other options are wrong

• (2) No single slip gives 32×10⁻³² J (eight times the correct value). Sanity check: (Q²)/(2C) = (64×10⁻³⁶)/(1.6×10⁻³) = 4×10⁻³² J.
• (3) No single slip gives 3.1×10⁻²⁶ J, nearly a million times too large. Sanity check: such a tiny charge on such a large capacitor stores almost no energy, (Q²)/(2C) = 4×10⁻³² J.
• (4) Puts the factor 2 on the wrong side: (2Q²)/C = (2×64×10⁻³⁶)/(8×10⁻⁴) = 16×10⁻³² J. The work is (Q²)/(2C), four times smaller.

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.