Question

Suppose that you charges a $\mathbf{\text{2.5 F}}$capacitor with two $\mathbf{\text{1.5 V}}$batteries. How much charge would be on each plate in the final state? How many excess electrons would be on the negative plate?

Short answer

The number of electrons on the negative plate is $4.68\times {10}^{19}\text{electrons}$.

Step-by-step solution

A concept:

When a capacitor is fully charged, there is a potential difference (p.d.) between its plates, and the larger the area of the plates and/or the smaller the distance between them (known as the separation), the greater the charge on the capacitor can accommodate and the larger its capacity will be.

Divide the entire excess charge from one electron's known charge and that gives the number of charges.

Given data:

The capacitor is charged with two $1.5\text{V}$ batteries, so the total voltage applied to the capacitor’s plates is,

$\begin{array}{rcl}V& =& 1.5\text{V+}1.5\text{V}\\ & =& 3\text{V}\end{array}$

The capacitance of the capacitor, $C=2.5\text{F}$

The magnitude of the charge on an electron,$e=1.6\times {10}^{-19}\text{C}$

Number of electrons on the negative plate:

Charge on the capacitor’s plate is given by a relation:

$Q=CV$ ..... (1)

The number of excess electrons on either of the plates is given by a relation,

$n=\frac{Q}{e}$ ..... (2)

Here, $Q$is the charge on the capacitor plate, $V$is the voltage applied to the capacitor plates, and $e$is the charge of the electron.

Using equation (1), the magnitude of the charge on each plate of the capacitor can be calculated as,

$\begin{array}{rcl}Q& =& CV\\ & =& \left(2.5\text{F}\right)\left(3.0\text{V}\right)\\ & =& 7.5\text{C}\end{array}$

Using equation (2), the number of excess electrons on the negative plate can be calculated as,

$\begin{array}{rcl}n& =& \frac{Q}{e}\\ & =& \frac{7.5\text{C}}{1.6\times {10}^{-19}\text{C}}\\ & =& 4.68\times {10}^{19}\text{electrons}\end{array}$

Hence, the number of electrons on the negative plate is $4.68\times {10}^{19}\text{electrons}$.