Question

Suppose that you charge a 1 F capacitor in a circuit containing two 1.5 V batteries, so the final potential difference across the plates is 3 V. How much charge is on each plate? How many excess electrons are on the negative plate?

Solutions

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Given

We are given two batteries connected to the capacitor where the emf of one battery is 1.5 V. So, the two batteries in series will produce emf = 2(1.5 V) = 3 V.

Solution

The capacitor is charged with two batteries 3 V\mathrm{V}, this means the potential difference between the two parallel plates is ΔV=3V\Delta V =3 \mathrm{V}. Where the potential difference is related to the capacitance and the charge on the plates by

C=QΔVQ=CΔV\begin{gather*} C = \dfrac{Q}{\Delta V}\\ Q=C \Delta V \tag{1} \end{gather*}

Now we can plug our values for CC and ΔV\Delta V into equation (1) to get the magnitude of the charge QQ on each plate

Q=CΔV=(1 F)(3 V)=3 C\begin{align*} Q &=C \Delta V \\ &=\left(1 \mathrm{~F}\right)(3 \mathrm{~V}) \\ &= \boxed{3 \mathrm{~C}} \end{align*}

The charges QQ on the plates is due to the number of electrons or positive charges. So, the number of excess electrons nn equals the charge QQ per the charge of one electron e=1.6×1019Ce = 1.6 \times 10^{-19} \,\text{C}

n=Qen=\frac{Q}{e}

Now we can plug our values for ee and QQ to get the number of excess electrons

n=Qe=3C1.6×1019C/electron=1.87×1019electrons\begin{align*} n = \frac{Q}{e} =\frac{3 \,\text{C}}{1.6 \times 10^{-19} \,\text{C/electron}} = \boxed{1.87\times 10^{19} \,\text{electrons}} \end{align*}

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