A capacitor with a capacitance of 100 is connected for 4.0 s across...

Time constant of the circuit is given by:

$\tau = RC = (20 \times 10^{3}) \times (100 \times 10^{-6}) = 2.0\text{ s}$

Maximum charge on the capacitor is:

$Q_{0} = CV = (100\mu\text{F}) \times (6.0\text{ V}) = 600\mu\text{C}$

During the charging phase for $t_{1} = 4.0\text{ s}$, the charge on the capacitor is:

$Q_{1} = Q_{0}(1 - e^{-t_{1}/\tau}) = Q_{0}(1 - e^{-4.0/2.0}) = Q_{0}(1 - e^{-2})$

When the battery is replaced by a thick wire, the capacitor discharges through the same resistor. The charge after discharging for $t_{2} = 4.0\text{ s}$ is:

$Q_{2} = Q_{1}e^{-t_{2}/\tau} = Q_{1}e^{-4.0/2.0} = Q_{1}e^{-2}$

Substituting the value of $Q_{1}$:

$Q_{2} = Q_{0}(1 - e^{-2})e^{-2} = Q_{0}(e^{-2} - e^{-4})$

Using $e^{-2} \approx 0.1353$ and $e^{-4} \approx 0.0183$:

$Q_{2} = 600(0.1353 - 0.0183) = 600(0.117) = 70.2\mu\text{C}$

The charge on the capacitor is approximately $70\mu\text{C}$.