91Ó°ÊÓ

The time constant, denoted as \(\tau\), is a crucial parameter in an RC circuit, representing the time it takes for the system to significantly react to a change in voltage. Specifically, \(\tau\) is the time required for the voltage across the capacitor to reach approximately 63.2% of its maximum value when charging, or decay to about 36.8% during discharging. It's calculated using the formula \(\tau = R \times C\), where \(R\) is the resistance and \(C\) is the capacitance.

• A larger time constant means the capacitor charges or discharges more slowly.
• Shorter time constants result in quicker changes in charge.

In this exercise, the resistance \(R\) is \(1.40 \times 10^6 \, \Omega\) and the capacitance \(C\) is \(1.80 \times 10^{-6} \, F\). By multiplying these values, we find \(\tau = 2.52 \, s\). This indicates the time scale over which the capacitor both approaches its maximum charge and loses that charge in an RC circuit.

Capacitance, symbolized by \(C\), refers to the ability of a capacitor to store electric charge. It is directly proportional to the amount of charge that can be held at a particular voltage. The unit of capacitance is the farad (F), but in practical applications, you'll often see microfarads (\(\mu\)F), nanofarads (nF), or picofarads (pF).

• A large capacitance signifies a high ability to store charge.
• Capacitance is determined by the size of the plates in the capacitor and the distance between them.

In our problem, the capacitor has a capacitance \(C = 1.80 \mu F\). This tells us how much electric charge it can store per volt of electric potential applied across its plates. The larger the capacitance, the larger the amount of charge at a given voltage.

The maximum charge \(Q_{max}\) that a capacitor can hold is defined by the equation \(Q_{max} = C \times \mathscr{E}\), where \(C\) represents the capacitance and \(\mathscr{E}\) is the applied emf or voltage. This expression provides insight into how much charge a capacitor can go up to as it fully charges.

• An increased emf results in increased maximum charge assuming capacitance stays constant.
• Understanding this helps design circuits needing specific charge storage.

For this exercise, with a capacitance \(C = 1.80 \times 10^{-6} \, \text{F}\) and voltage \(\mathscr{E} = 12.0 \, \text{V}\), the maximum charge is determined as \(Q_{max} = 21.6 \, \mu C\). This is how much charge the capacitor will hold once it's fully charged under this circuit setup.