# Potential at a point due to two charges

• 6Stang7
In summary, the potential at a point on the z-axis above the x-axis is given by: r = 2*cos(theta) where theta is the angle between the z-axis and the x-axis.
6Stang7
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:

The potential due to a charge is given by:

We can express r as:

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:

This gives us:

Putting everything together we get:

Therefore, the total potential at point P is:

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

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

Which when solved give:

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

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

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

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.

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.

Dear student,

Thank you for sharing your work and thought process on this problem. It seems like you have a good understanding of the concept and have made a logical attempt at solving it.

After reviewing your work, I believe the mistake lies in the expression for the z-component of the potential. You have correctly identified that the x-components will cancel out due to symmetry, but you have used the cosine of the magnitude instead of the cosine of the angle between the z-axis and the vector connecting the charge to point P. This angle is not equal to the magnitude of the vector, so using the cosine of the magnitude will give an incorrect result.

The correct expression for the z-component of the potential should be:

Vz = (kq/d) * cos(θ)

Where θ is the angle between the z-axis and the vector connecting the charge to point P. This can be calculated using trigonometry as θ = tan^-1(z/d).

When you plug this into your final equation for the potential, you should get the correct result of:

V = (2kq/d) * (1 - z/d)

I hope this helps to clarify the mistake in your calculation. Keep up the good work!

Best regards,

## 1. What is potential at a point due to two charges?

Potential at a point due to two charges is a measure of the electric potential energy at a specific point in space, caused by the presence of two charges. It is affected by the magnitude and distance of the charges.

## 2. How is potential at a point due to two charges calculated?

The potential at a point due to two charges can be calculated using the formula V = k(q1/r1 + q2/r2), where V is the potential, k is the Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r1 and r2 are the distances from the point to each charge.

## 3. What is the unit of potential at a point?

The unit of potential at a point is volts (V), which is equivalent to joules per coulomb (J/C).

## 4. How does the potential at a point due to two charges change with distance?

The potential at a point due to two charges decreases with distance. As the distance from the charges increases, the potential decreases because the electric field becomes weaker.

## 5. Can potential at a point due to two charges be negative?

Yes, the potential at a point due to two charges can be negative. This indicates that the potential energy at that point is negative, meaning that it requires work to move a test charge from infinity to that point against the electric field.

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