A capacitor with capacitance C = 6.00 μF is fully charged by a 50.0 V battery. After being disconnected from the battery, the capacitor discharges through a resistor with resistance R = 185 Ω. What is the charge q on the capacitor when the current in the resistor is 0.180 A?

To find the charge q on the capacitor when the current in the resistor is 0.180 A, we first need to know the relationship between the voltage across the resistor, the resistance, and the current through it, given by Ohm's Law:

V = I × R

where V = voltage across the resistor I = current through the resistor R = resistance

Given the current I = 0.180 A and resistance R = 185 Ω, we can calculate the voltage V across the resistor:

V = I × R = 0.180 A × 185 Ω = 33.3 V

Since the capacitor is initially charged by a 50.0 V battery and then allowed to discharge through the resistor, the voltage across the capacitor (Vc) has decreased to 33.3 V at this instant because the rest of the voltage drop occurs across the resistor. Therefore, Vc = 33.3 V.

Next, we use the relation between the charge q on a capacitor and its voltage Vc, given by the formula for capacitance:

q = C × Vc

where C = capacitance of the capacitor Vc = voltage across the capacitor

Substituting the values of C = 6.00 μF (or 6.00 × 10^-6 F) and Vc = 33.3 V, we get:

q = 6.00 × 10^-6 F × 33.3 V = 1.998 × 10^-4 C, or q ≈ 200 μC.

So, the charge on the capacitor when the current in the resistor is 0.180 A is approximately 200 μC.