184_notes:examples:week2_electric_potential_negative_point

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184_notes:examples:week2_electric_potential_negative_point [2017/08/28 21:35] – [Solution] tallpaul184_notes:examples:week2_electric_potential_negative_point [2018/05/17 16:49] (current) curdemma
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 +[[184_notes:pc_potential|Return to Electric Potential]]
 ===== Example: Electric Potential from a Negatively Charged Balloon ===== ===== 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 at a point $P$, which is a distance $R=20 \text{ m}$ from the center of mass of the balloon? A diagram below shows a rough sketch. +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?
-{{ 184_notes:2_potential_positive_balloon.png?150 |Charged Balloon, and Point P}}+
  
 ===Facts=== ===Facts===
   * The balloon has total charge $Q=-5.0\cdot 10^{-9} \text{ C}$.   * 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 mass of the balloon. +  * 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 $$\frac{1}{4\pi\epsilon_0}\frac{q}{r},$$ where $qrepresents the charge and $ris the distance.
-===Lacking=== +
-  * The electric potential at $P$+
- +
-===Approximations & Assumptions=== +
-  * The electric potential at $Pis due entirely to the balloon. +
-  * $Plies outside of the balloon. +
-  * The balloon's electric field outside the balloon acts like a point charge centered at the center of mass of the balloon. +
-  * The electric potential infinitely far away from the balloon is $0 \text{ V}$.+
  
 ===Representations=== ===Representations===
-  * The electric potential due to a point charge (to which we are approximating the balloon) can be written as $$= \frac{1}{4\pi\epsilon_0}\frac{q}{r},$$ where $q$ represents our charge and $r$ is our distance.+<WRAP TIP> 
 +=== Assumption === 
 +We assume $P$ lies outside of the balloon. This is obvious, as $Pis a distance $R=20 \textm}$ 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==== ====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 The electric potential at $P$ is given by
 \begin{align*} \begin{align*}
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   &= -2.2 \text{ V}   &= -2.2 \text{ V}
 \end{align*} \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 the point charge, the higher the electric potential]].+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 point charge, the higher the magnitude of electric potential]].
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