Potential at a point due to two charges

6Stang7 · Nov 7, 2010

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:

Diagram of two +q charges on the horizontal axis separated by distance d, with a point above the midpoint at height z, showing distance r and angle φ to the upper point.

The potential due to a charge is given by:

$Equation for electric potential of a point charge: V = (1 / (4 pi epsilon_0)) * (q / r), with the fraction written as a typeset formula.$

We can express r as:

Equation r = square root of (z squared plus (d/2) squared), written as a formula for the distance from a point to a charge.

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:

Two equations for the potential V of a point charge: V = (1/(4πε0)) (q/r) cos φ, which equals (1/(4πε0)) q z / r^2.

Putting everything together we get:

Equation for potential V on the z-axis from two charges: V = (1/(4 pi epsilon_0)) times qz over the square root of (sqrt(z^2 + (d/2)^2)) squared, equals (1/(4 pi epsilon_0)) times qz over (z^2 + (d/2)^2) squared.

Therefore, the total potential at point P is:

Equation V = (1 / (4 pi epsilon_0)) times (2qz / (z^2 + (d/2)^2)), with the second factor's denominator written as z squared plus (d over 2) squared.

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):

Equation: negative gradient of V equals the partial derivative with respect to z of 1/(4 pi epsilon_0) times 2qz/(z^2 + (d/2)^2).

Which when solved give:

$Equation for the electric field E = (2q / 4πε₀) times a fraction: numerator ((d/2)² − z²), denominator (z² + (d/2)²) squared.$

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

Handwritten-style derivation of an electric field on the axis of a point charge: E = 1/(4πε0) q/r², r = √(z² + (d/2)²), cos φ = z/r, E = 1/(4πε0) qz/r³, and E = 1/(4πε0) qz/(z² + (d/2)²)^(3/2).

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

Formula for the electric field E = (1/(4πε₀)) · 2qz / (z² + (d/2)²)^(3/2), shown on a yellow background.

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

Equation for the potential V at a point on the z-axis: V = (1/(4πε0)) · 2q / sqrt(z² + (d/2)²).

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.

stevenb · Nov 7, 2010

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.

6Stang7 · Nov 7, 2010

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.

stevenb · Nov 8, 2010

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.

Potential at a point due to two charges

Homework Statement

The Attempt at a Solution

Discussion

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