More Charge Examples

Two identical, insulated spheres have different charges. Sphere 1 has a charge of \(-\text{96} \times \text{10}^{-\text{18}}\) \(\text{C}\). Sphere 2 has \(\text{60}\) excess electrons. If the two spheres are brought into contact and then separated, what charge will each have?

Step 1: Analyze 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.

Two identical, metal spheres on insulating stands have different charges. Sphere 1 has a charge of \(-\text{96} \times \text{10}^{-\text{18}}\) \(\text{C}\). Sphere 2 has \(\text{60}\) excess protons. If the two spheres are brought into contact and then separated, what charge will each have? How many electrons or protons does this correspond to?

Step 1: Analyze 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.

Step 2: 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 \(\text{1.6} \times \text{10}^{-\text{19}}\) \(\text{C}\)

Step 3: 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.

Two identical, metal spheres have different charges. Sphere 1 has a charge of \(-\text{9.6} \times \text{10}^{-\text{18}}\) \(\text{C}\). Sphere 2 has \(\text{30}\) excess electrons. If the two spheres are brought into contact and then separated, what charge will each have? How many electrons does this correspond to?

Step 1: Analyze 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.

Step 2: 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*}

Step 3: 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 \(-\text{1.6} \times \text{10}^{-\text{19}}\) \(\text{C}\).

Step 4: 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*}