A capacitor with a capacitance of 100 F is connected for 4.0 s across a 6.0 V battery through a 20 k resistance. Following this, the battery is replaced by a thick wire. Determine the charge on the capacitor 4.0 s after disconnecting the battery.

Step-by-step solution

Time constant of the circuit is given by: = RC = (20 10³) (100 10⁻⁶) = 2.0 s Maximum charge on the capacitor is: Q₀ = CV = (100 F ) (6.0 V ) = 600 C During the charging phase for t₁ = 4.0 s , the charge on the capacitor is: Q₁ = Q₀(1 - e^ -t₁/ ) = Q₀(1 - e^ -4.0/2.0 ) = Q₀(1 - e⁻²) When the battery is replaced by a thick wire, the capacitor discharges through the same resistor. The charge after discharging for t₂ = 4.0 s is: Q₂ = Q₁e^ -t₂/ = Q₁e^ -4.0/2.0 = Q₁e⁻² Substituting the value of Q₁ : Q₂ = Q₀(1 - e⁻²)e⁻² =