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--- 184_notes:examples:week2_electric_potential_negative_point [2017/08/25 19:44] – created tallpaul
+++ 184_notes:examples:week2_electric_potential_negative_point [2018/05/17 16:49] (current) – curdemma
@@ Line 1: / Line 1: @@
-yeas
+[[184_notes:pc_potential|Return to Electric Potential]]
+===== Example: Electric Potential from a Negatively Charged Balloon =====
+Suppose we have a negatively 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  $R=20 \text{ m}$  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  $R=20 \text{ m}$  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. This is obvious, as  $P$  is a distance  $R=20 \text{ m}$  away from the center of the balloon.
+</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 a point very far from the balloon, so the balloon would "look" like a point charge from the perspective of an observation location that is  $20 \text{ m}$  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{-5.0\cdot 10^{-9} \text{ C}}{20 \text{ m}} \\
+  &= -2.2 \text{ V}
+\end{align*}
+Notice how the magnitude of charge on the balloon is the same as in the "positively charged balloon" [[184_notes:examples:Week2_electric_potential_positive_point|example]]. The reason the magnitude of the voltage is so much smaller, is because the distance is so much greater. [[184_notes:pc_potential#Potential_vs_Distance_Graphs|The closer you get to a point charge, the higher the magnitude of electric potential]].