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December 11th, 2017, 03:50 PM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 40 Thanks: 1 Math Focus: Calculus  How to solve this problem related to electric potential?
The problem is as follows: At 1 meter to the left of a particle whose charge is $\displaystyle q_{1}=1.0 \mu C$ there exists another whose charge is $\displaystyle q_{2}=1.0 \mu C$. Find the electric potential resulting from both point charges located at $\displaystyle \textrm{1.0 m}$ to the right of the particle $\displaystyle q_{1}$. So far I found that the equation that should I use is this one: $\displaystyle v= k \times \frac{kq}{r}$ But I'm confused on how should I use it. I thought since the distance which separates both is 1 meter only applies for the electric potential of $q_{1}$ but there is an additional 1 meter for the other charge $q_{2}$ as the place which the problem states the electric potential should be calculated is at one meter to the right of $q_{1}$. But I am not sure if the way how I translated those facts into the above equation are correct?. Would it be like this? $\displaystyle v= k \times \frac{kq}{r}$ $\displaystyle v_{total}= v_{1} + v_{2}$ $\displaystyle v= k \times \frac{kq_{1}}{r_{1}} + k \times \frac{kq_{2}}{r_{2}}$ Plugin the values from the problem: $\displaystyle v= (9.0 \times 10^{9} Nm^{2}C^{2}) \times \frac{+1.0\times 10^{6}C}{1 m} + (9.0 \times 10^{9} Nm^{2}C^{2}) \times \frac{1.0\times 10^{6}C}{2 m}$ $\displaystyle v= (9.0 \times 10^{9} Nm^{2}C^{2}) \times 10^{6}C (\frac{21}{2m})$ $\displaystyle v= (9.0 \times 10^{9} Nm^{2}C^{2}) \times 10^{6}C (\frac{1}{2m})= 4.5 \times 10^{3} Nm^{1}C^{1}) \textrm{ or } \frac{J}{C} \textrm{or Volt}$ I'm adding a sketch of how I thought this problem. 
December 11th, 2017, 04:19 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,755 Thanks: 1405  Quote:
since voltage is a scalar quantity ... $V = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}$ $V = k\left(\dfrac{1 \, \mu C}{2 \, m} + \dfrac{1 \, \mu C}{1 \, m}\right)$ $V = \left(9 \times 10^9 \, \dfrac{Nm^2}{C^2} \right)\left(\dfrac{1 \, \mu C}{2 \, m} \right)$ $V = 4.5 \times 10^3 \, \dfrac{Nm}{C}$ check out the link ... Electric potential for multiple point charges  
December 11th, 2017, 05:44 PM  #3  
Member Joined: Jun 2017 From: Lima, Peru Posts: 40 Thanks: 1 Math Focus: Calculus  Quote:
This was another question I had when solving the problem, whether the contributions of each charge only indicate whether is positive or negative of its numeric value and does not take into account the direction. Moreover I thought that if both charges have different polarities one would attract the other and as a result the total distance would become zero. Which would make the fraction of $\frac{q}{r}$ impossible to determine. It seems that such conclusion was incorrect. Quote:  

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electric, electric potential, point charge, potential, problem, related, solve 
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