Final answer:
The capacitance of the capacitor is determined by dividing the charge on the capacitor by the voltage across it. In this case, the given voltage is 47 volts. So, the capacitance of the capacitor is 2.5 microfarads.
Step-by-step explanation:
The student's question asks about determining the capacitance of a capacitor when the voltage across its plates is given. However, the question initially mentions a '2.5 capacitor' without units, which seems to be a mistake. Assuming the student meant to say '2.5 μF' (microfarads), the capacitance of the capacitor is already given as 2.5 μF. Therefore, no additional calculation is required to determine the capacitance since it has been stated in the question. If the student meant to find something else, such as the charge stored on the capacitor or the energy stored, they would need to use the appropriate formulas: Q = C × V for charge (where Q is charge, C is capacitance, and V is voltage) and \(U = \frac{1}{2} C V^2\) for energy (where U is energy, C is capacitance, and V is voltage).
The capacitance of a capacitor is determined by dividing the charge on the capacitor by the voltage across it. In this case, the given voltage is 47 volts and the charge is not provided. So, we need to find the charge first. The charge is given by multiplying the capacitance and the voltage. Therefore, the charge on the capacitor is given by:
Charge = Capacitance * Voltage = 2.5 * 47 = 117.5
Now, we can determine the capacitance by dividing the charge by the voltage:
Capacitance = Charge / Voltage = 117.5 / 47 = 2.5
Therefore, the capacitance of the capacitor is 2.5 microfarads.