[[184_notes:pc_potential|Return to Electric Potential]]
===== Example: Electric Potential from a Positively Charged Balloon =====
Suppose we have a positively charged balloon with total charge $Q=5.0\cdot 10^{-9} \text{ C}$. What is the electric potential (also called voltage) at a point $P$, which is a distance $R=50 \text{ cm}$ from the center of the balloon?

===Facts===
  * The balloon has total charge $Q=5.0\cdot 10^{-9} \text{ C}$.
  * The point $P$ is a distance $R=50 \text{ cm}$ away from the center of the balloon.
  * The electric potential due to a point charge can be written as $$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r},$$ where $q$ represents the charge and $r$ is the distance.

===Representations===
<WRAP TIP>
=== Assumption ===
We assume $P$ lies outside of the balloon. We make this assumption because it was not specified, but this seems to make more sense than $P$ being inside the balloon. This also helps us draw the representation below, which can be used to bolster our approximation later on of the balloon as a point charge.
</WRAP>

[{{ 184_notes:2_potential_positive_balloon.png?150 |Charged Balloon, and Point P}}]

===Goal===
  * Find the electric potential at $P$.

====Solution====
<WRAP TIP>
=== Approximation ===
We approximate the balloon as a point charge. We do this because we have the tools to find the electric potential from a point charge. This seems like a reasonable approximation because the balloon is not too spread out, and we are interested in points outside the balloon. so the balloon might "look" like a point charge from the perspective of an observation location that is little far away.
</WRAP>

<WRAP TIP>
=== Assumption ===
The electric potential infinitely far away from the balloon is $0 \text{ V}$. Read [[184_notes:superposition#Superposition_of_Electric_Potential|here]] for why this is important.
</WRAP>

The electric potential at $P$ is given by
\begin{align*}
V &= \frac{1}{4\pi\epsilon_0}\frac{q}{r} \\
  &= \frac{1}{4\pi\cdot 8.85\cdot 10^{-12} \frac{\text{C}}{\text{Vm}}}\frac{5\cdot 10^{-9} \text{ C}}{0.5 \text{ m}} \\
  &= 90 \text{ V}
\end{align*}