Potential at a point due to two charges

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SUMMARY

The discussion centers on calculating the electric potential at a point P above two positive charges, each valued at +q, separated by a distance d on the x-axis. The user initially miscalculates the potential by incorrectly applying the cosine function to determine the z-component of the potential. It is clarified that electric potential is a scalar quantity and should not be treated as a vector. The correct approach emphasizes that potential does not have direction, and the confusion arises when considering the effects of negative charges on the potential.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with scalar and vector quantities in physics
  • Knowledge of Coulomb's Law and its application
  • Basic trigonometry, specifically the cosine function
NEXT STEPS
  • Review the concept of electric potential and its scalar nature
  • Study the relationship between electric potential and electric field
  • Learn about the effects of negative charges on electric potential
  • Explore the implications of conservative forces in electric fields
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric potential and fields in electrostatics.

6Stang7
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I've been working on this problem and I cannot find out where I am making a mistake.

Homework Statement


Two charges, each with a value of +q, are placed a distance d apart on the x-axis. Find the potential at a point P a distance z above the x-axis on the z-axis

The Attempt at a Solution



The figure below shows the physics system:

eqoea.jpg


The potential due to a charge is given by:

Smthy.jpg


We can express r as:

Y8Sg8.jpg


We can see by symmetry that the x-components will cancel out, so the total potential will be twice of the z-component. The z-component of the potential is the cosine of the magnitude, where cosine is:

KOcsg.jpg


This gives us:

keUqf.jpg


Putting everything together we get:

Wr8Up.jpg


Therefore, the total potential at point P is:

o9Ha1.jpg


Now, we know that the potential is related to the Electric Field by:

o8TZL.jpg


This means that the Electric Field at the same point is (I realize the forget the minus sign just now):

KvB52.jpg


Which when solved give:

Jh1ok.jpg


Now, using the same figure, but instead of solving for the potential, we solve for the Electric Field, we find:

FwTtY.png


The above uses the same line of thinking, and this gives a total value of the Electric Field to be:

LWKm6.png


The problem is that the book says that the potential I calculated above is wrong, and that the value should be:

0yGMN.jpg


When you calculate the value of the Electric Field from this, you get the same answer as the value I just calculated.

So, what did I do wrong in my calculation of the potential? The error seems to steam from the part where I calculated the z-component of one of the charges using cosine, but I can't see why that is incorrect.
 
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6Stang7 said:
I've been working on this problem and I cannot find out where I am making a mistake.So, what did I do wrong in my calculation of the potential? The error seems to steam from the part where I calculated the z-component of one of the charges using cosine, but I can't see why that is incorrect.

Just remember that potential is not a vector, but is a scalar. You don't add scalars vectorially.
 
So if one of the charges was negative instead of positive, the potential would be zero?

What confuses me is that if one charge was changed to negative, then there'd be a constant potential in the x-direction, but zero in the z-direction.
 
6Stang7 said:
So if one of the charges was negative instead of positive, the potential would be zero?

Yes, on the z axis it would be.

6Stang7 said:
What confuses me is that if one charge was changed to negative, then there'd be a constant potential in the x-direction, but zero in the z-direction.
Remember, potential does not have a direction since it is a scalar and not a vector. There will be a force, or equivalently an electric field (which is force per unit charge) in the direction you say.

If potential is referenced to infinity, then the work done to bring a charge from infinity to any finite point on the z axis is clearly zero. Why, because you can chose the z axis itself as the path, and it's clear that the dot product of the electric field and the path is always zero. The x-component of electric field (and force) is orthogonal to the chosen path. The fact that electric fields are conservative tells you that any other path will yield the same value.
 

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