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

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

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

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

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

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

Now we can plug our values for $e$ and $Q$ to get the number of excess electrons

\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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