Electric Potential Energy of Doorknob and Dust Speck

Consider again the doorknob from the example in the previous section. The doorknob is treated as a spherical conductor with a uniform static charge $q_{1} = −1.5 nC$ on its surface. What is the electric potential energy between the doorknob and a speck of dust carrying a charge $q_{2} = 0.20 nC$ at 1.0 cm from the front surface of the doorknob? The diameter of the doorknob is 5.0 cm.

This diagram shows the edge of a door in cross-section and a doorknob attached to it. Two axes, perpendicular to each other, originate from the center of the doorknob and point away from it. The horizontal axis is labeled “plus x circumflexΣ and the vertical axis is labeled “plus y circumflex”. A dot on the horizontal axis beyond the doorknob is labeled “Speck of dust”, and a second label that gives the equation “x = 3.5 cm”.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
As we did in the previous section, we treat the charge as if it were concentrated at the center of the doorknob. Again, as you will be able to validate in later physics classes, we can make this simplification, because the charge is uniformly distributed over the surface of the spherical object. Make a sketch of the situation and define a coordinate system, as shown in the image below. We use $+ x$ to indicate the outward direction perpendicular to the door, with $x = 0$ at the center of the doorknob. If the diameter of the doorknob is 5.0 cm, its radius is 2.5 cm. Thus, the speck of dust 1.0 cm from the surface of the doorknob is a distance $r = 2.5 cm + 1.0 cm = 3.5 cm$ from the center of the doorknob. To solve this problem, use the equation $U_{E} = k q_{1} q_{2} / r$ .
Solution
The charge on the doorknob is $q_{1} = −1.5 nC = −1.5 \times 10^{−9} C,$ and the charge on the speck of dust is $q_{2} = 0.20 nC = 2.0 \times 10^{−10} C$ . The distance $r = 3.5 cm = 0.035 m$ . Inserting these values into the equation $U_{E} = k q_{1} q_{2} / r$ gives

$\begin{matrix} U_{E} & = & \frac{k q_{1} q_{2}}{r} \\ = & \frac{(8.99 \times 10^{9} N \cdot m^{2} {/C}^{2}) (- 1.5 \times 10^{−9} C) (2.0 \times 10^{−10} C)}{(0.035 m)} \\ = & −7.7 \times 10^{−8} J . \end{matrix}$
18.34

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