Question

A 1.0-mm-diameter ball bearing has $2.0\times 109$excess electrons.

What is the ball bearing’s potential?

Short answer

The potential energy of the ball bearing is $-5.76\mathrm{kV}$

Step-by-step solution

Concept of  the potential energy of the ball bearing 

In order to find the electric energy $U$the electric potential $V$of the sphere needs to be found out.

Electrostatic Potential of $V$is directly proportional to the charge $q$and inversely proportional to the distance from the point of charge. This statement is mathematically expressed as

$V=k\frac{q}{r}$

In the expression above, $\left(k\right)$is the Coulomb's Constant.

role="math" localid="1648204492509" $k=9\times {10}^9{\text{N.m}}^2/{\mathrm{C}}^2$

Finding the excess charge on the Ball Bearing

The expression of Excess Charge on the Ball Bearing can be represented as;

$q=ne$

As per data provided;

Substitute $n$with $2\times {10}^9$and $e$with $-1.6\times {10}^{-19}C$in the equation above.

Therefore,

role="math" localid="1648204858333" $\begin{array}{rcl}q& =& ne\\ & =& (2\times {10}^9)(-1.6\times {10}^{-19}C)\\ & =& -3.2\times {10}^{-10}C\end{array}$

Finding Radius of the Ball Bearing

Converting the unit of diameter of the Ball Bearing from $mm$to $m$

role="math" localid="1648205103240" $\begin{array}{rcl}d& =& 1.0\mathrm{mm}\\ & =& (1.0\mathrm{mm})\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)\\ & =& 1.0\times {10}^{-3}\mathrm{m}\\ & & \end{array}$

Therefore, radius of the ball bearing is,

$\begin{array}{rcl}r& =& \frac{d}{2}\\ & =& \frac{1.0\times {10}^{-3}\mathrm{m}}{2}\\ & =& 0.5\times {10}^{-3}\mathrm{m}\end{array}$

Finding the Ball Bearing's Potential

In order to solve this problem, approximation of Point Charge has to be taken into Consideration. Since the Ball Bearing is a Conducting Sphere, the potential of the sphere in the centre is equal to the potential at the surface.

Calculation of the the potential due to point charge at a distance $\text{'}r\text{'}$can be written as

localid="1648205847436" $V=\frac{kq}{r}$

We now need to substitute the values of $k$, $q$and $r$.

$k=9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$

$q=-3.2\times {10}^{-10}\mathrm{C}$

role="math" localid="1648205799784" $r=0.5\times {10}^{-3}\mathrm{m}$

$\begin{array}{rcl}\mathrm{V}& =& \frac{\mathit{k}\mathit{q}}{\mathit{r}}\\ & =& \frac{(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2)(-3.2\times {10}^{-10}\mathrm{C})}{(0.5\times {10}^{-3}\mathrm{m})}\\ & =& -\left(5760\mathrm{V}\right)\left(\frac{1\mathrm{kV}}{{10}^3\mathrm{V}}\right)\\ & =& -5.76\mathrm{kV}\end{array}$

Thus, the potential difference isrole="math" localid="1648206107263" $-5.76\mathrm{kV}$