Problem with solution: Charge, Capacitance and Half-Life of a Capacitor When Discharging From 50V to 15V.

To find out the capacitance \(C\) of the capacitor, we use the formula that describes the time dependence of the voltage across the capacitor during the discharge process:

What we are looking for is the capacitance, so we rearrange Eq. 1 for \(C\):

The voltage \( U_{\mathrm C}(1\,\mathrm{ms}) = 15 \, \mathrm{V} \) after \( t = 1 \,\mathrm{ms} \) is given. The initial voltage \( U_0 = 50 \, \mathrm{V} \) and in series connected resistor \(R = 2000 \mathrm{\Omega} \) are also known. Inserting the concrete values yields the capacitance we are looking for:

The time constant \( \tau \) is given by:

It describes the time after which the initial voltage has dropped to about 37%. Inserting the discharge resistance and the capacitance found in a) yields:

In this exercise we want to calculate the half-life \( t_{\mathrm h} \). The half-life indicates the time after which the initial voltage of \( U_0 = 50 \, \mathrm{V} \) has dropped to half, that is to \( U_{\mathrm C} = \frac{U_0}{2} = 25 \, \mathrm{V} \):

Rearranging for half-life \( t_{\mathrm h} \) yields:

As you can see, the half-life is determined by the time constant \(R \, C \), which is simply multiplied by \( \ln(2) \). So we can simply use the time constant calculated in b) for the half-life:

Here we want to find out how much charge \(Q_0\) was on the capacitor before the capacitor was discharged. So we are looking for capacitor charge at the initial time \(t = 0\). To do this, we use the relationship between the voltage and charge on the capacitor:

Of course we need the initial voltage \( U_0 \), because this was applied at the initial time \(t = 0\). Inserting the values results in: