A capacitor with capacitance 5 μ F is charged to 5 μC. If the plates are pulled apart to reduce the capacitance to 2 μ F, how much work is done?

Question

Accepted Answer

(2) 3.75×10⁻⁶ J. The plates are pulled apart with no battery attached, so the charge stays at Q=5 μC and U=(Q²)/(2C). Uᵢ=((5×10⁻⁶)²)/(2×5×10⁻⁶)=(25×10⁻¹²)/(10⁻⁵)=2.5×10⁻⁶ J. U_f=(25×10⁻¹²)/(2×2×10⁻⁶)=6.25×10⁻⁶ J. The work done in separating the plates is the rise in stored energy: W=6.25-2.5=3.75×10⁻⁶ J.

A capacitor with capacitance 5 μ F is charged to 5 μC. If the plates are pulled apart to reduce the capacitance to 2 μ F, how much work is done?

Step-by-step solution

Why the other options are wrong