16.4 Quantisation of charge

Analyse the problem and identify the principles

We are asked to determine a number of electrons based on a total charge. We know that charge is quantised and that electrons carry the base unit of charge which is \(-\text{1,6} \times \text{10}^{-\text{19}}\) \(\text{C}\).

Apply the principle

As each electron carries the same charge the total charge must be made up of a certain number of electrons. To determine how many electrons we divide the total charge by the charge on a single electron:

\begin{align*} N & = \frac{-\text{1,92} \times \text{10}^{-\text{17}}}{-\text{1,6} \times \text{10}^{-\text{19}}} \\ & = 120 \text{ electrons} \end{align*}

Analyse the question

We have two identical negatively charged conducting spheres which are brought together to touch each other and then taken apart again. We need to explain what happens to the charge on each sphere and what the final charge on each sphere is after they are moved apart.

Identify the principles involved

We know that the charge carriers in conductors are free to move around and that charge on a conductor spreads itself out on the surface of the conductor.

Apply the principles

1. When the two conducting spheres are brought together to touch, it is as though they become one single big conductor and the total charge of the two spheres spreads out across the whole surface of the touching spheres. When the spheres are moved apart again, each one is left with half of the total original charge.

2. Before the spheres touch, the total charge is: \(-\text{5}\text{ nC} + (-\text{3}\text{ nC}) = -\text{8}\text{ nC}\). When they touch they share out the \(-\text{8}\) \(\text{nC}\) across their whole surface. When they are removed from each other, each is left with half of the original charge:

   \[\frac{-\text{8}\text{ nC}}{2} = -\text{4}\text{ nC}\]

   on each sphere.

Analyse the question

We need to determine what will happen to the charge when the spheres touch. They are insulators so we know they will NOT allow charge to move freely. When they touch nothing will happen.

Analyse the question

We need to determine what will happen to the charge when the spheres touch. They are metal spheres so we know they will be conductors. This means that the charge is able to move so when they touch it is possible for the charge on each sphere to change. We know that charge will redistribute evenly across the two spheres because of the forces between the charges. We need to know the charge on each sphere, we have been given one.

Identify the principles involved

This problem is similar to the earlier worked example. This time we have to determine the total charge given a certain number of protons. We know that charge is quantised and that protons carry the base unit of charge and are positive so it is \(\text{1,6} \times \text{10}^{-\text{19}}\) \(\text{C}\)

Apply the principles

The total charge will therefore be:

\begin{align*} {Q}_{2} & = 60\times \text{1,6} \times \text{10}^{-\text{19}}\text{ C} \\ & = \text{9,6} \times \text{10}^{-\text{18}}\text{ C} \end{align*}

As the spheres are identical in material, size and shape the charge will redistribute across the two spheres so that it is shared evenly. Each sphere will have half of the total charge:

\begin{align*} Q& = \frac{{Q}_{1} + {Q}_{2}}{2} \\ & = \frac{\text{9,6} \times \text{10}^{-\text{18}}\text{ C} + \left(-\text{9,6} \times \text{10}^{-\text{18}}\text{ C}\right)}{2}\\ & = \text{0}\text{ C} \end{align*}

So each sphere is now neutral.

No net charge means that there is no excess of electrons or protons.

Analyse the problem

We need to determine what will happen to the charge when the spheres touch. They are metal spheres so we know they will be conductors. This means that the charge is able to move so when they touch it is possible for the charge on each sphere to change. We know that charge will redistribute evenly across the two spheres because of the forces between the charges. We need to know the charge on each sphere, we have been given one.

Identify the principles

This problem is similar to the earlier worked example. This time we have to determine the total charge given a certain number of electrons. We know that charge is quantised and that electrons carry the base unit of charge which is \(-\text{1,6} \times \text{10}^{-\text{19}}\) \(\text{C}\)

\begin{align*} {Q}_{2} & = 30 \times -\text{1,6} \times \text{10}^{-\text{19}}\text{ C}\\ & = -\text{4,8} \times \text{10}^{-\text{18}}\text{ C} \end{align*}

Apply the principles: redistributing charge

As the spheres are identical in material, size and shape the charge will redistribute across the two spheres so that it is shared evenly. Each sphere will have half of the total charge:

\begin{align*} Q & = \frac{{Q}_{1} + {Q}_{2}}{2} \\ & = \frac{-\text{9,6} \times \text{10}^{-\text{18}} + \left(-\text{4,8} \times \text{10}^{-\text{18}}\right)}{2} \\ & = -\text{7,2} \times \text{10}^{-\text{18}}\text{ C} \end{align*}

So each sphere now has \(-\text{7,2} \times \text{10}^{-\text{18}}\) \(\text{C}\) of charge.

We know that charge is quantised and that electrons carry the base unit of charge which is \(-\text{1,6} \times \text{10}^{-\text{19}}\) \(\text{C}\).

Apply the principles: charge quantisation

As each electron carries the same charge the total charge must be made up of a certain number of electrons. To determine how many electrons we divide the total charge by the charge on a single electron:

\begin{align*} N & = \frac{-\text{7,2} \times \text{10}^{-\text{18}}}{-\text{1,6} \times \text{10}^{-\text{19}}} \\ & = 45 \text{ electrons} \end{align*}