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December 11th, 2017, 03:50 PM   #1
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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?

$\displaystyle v= k \times \frac{kq}{r}$

$\displaystyle v_{total}= v_{1} + v_{2}$

$\displaystyle v= k \times \frac{kq_{1}}{r_{1}} + k \times \frac{kq_{2}}{r_{2}}$

Plug-in the values from the problem:

$\displaystyle v= (9.0 \times 10^{9} Nm^{2}C^{-2}) \times \frac{+1.0\times 10^{-6}C}{1 m} + (9.0 \times 10^{9} Nm^{2}C^{-2}) \times \frac{-1.0\times 10^{-6}C}{2 m}$

$\displaystyle v= (9.0 \times 10^{9} Nm^{2}C^{-2}) \times 10^{-6}C (\frac{2-1}{2m})$

$\displaystyle v= (9.0 \times 10^{9} Nm^{2}C^{-2}) \times 10^{-6}C (\frac{1}{2m})= 4.5 \times 10^{3} Nm^{1}C^{-1}) \textrm{ or } \frac{J}{C} \textrm{or Volt}$

I'm adding a sketch of how I thought this problem.
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December 11th, 2017, 04:19 PM   #2
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Quote:
 Originally Posted by Chemist116 ... So far I found that the equation that should I use is this one: $\displaystyle v= k \times \frac{kq}{r}$
why do you have the Coulomb constant, $k$, in the equation twice? once is enough ... $V = \dfrac{kQ}{r}$

since voltage is a scalar quantity ...

$V = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}$

$V = k\left(\dfrac{-1 \, \mu C}{2 \, m} + \dfrac{1 \, \mu C}{1 \, m}\right)$

$V = \left(9 \times 10^9 \, \dfrac{Nm^2}{C^2} \right)\left(\dfrac{1 \, \mu C}{2 \, m} \right)$

$V = 4.5 \times 10^3 \, \dfrac{Nm}{C}$

Electric potential for multiple point charges

December 11th, 2017, 05:44 PM   #3
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From: Lima, Peru

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Math Focus: Calculus

Quote:
 Originally Posted by skeeter why do you have the Coulomb constant, $k$, in the equation twice? once is enough ... $V = \dfrac{kQ}{r}$
Sorry it was a typo during the process when I transcribed the method but as you can see I omitted the "double k" in my calculations.

Quote:
 Originally Posted by skeeter since voltage is a scalar quantity ...
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:
 Originally Posted by skeeter check out the link ... Electric potential for multiple point charges

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