Difference in potential between point charges

[nobrainer612]

Homework Statement

Below is a graphic representation of the problem.

With a point charge at the origin, calculate the difference in potential of point B relative to point A using the path integral.

Homework Equations

As I recalled, to find the electric potential difference, Vab = - ∫ E ° dl (° = dot product)
Potential difference for point charge: V = Q/ 4πεr
Vab = Vb - Va
Electric field for point charge should be, E = Q/4πε(R^2)

The Attempt at a Solution

I am not sure what is meant by path integral. But I will try to do it but not sure the correctness:
Since there is one point charge, and it is located at the origin. SO
V = -∫E°dl = -∫(Q/4πε(R^2) ) dr

V = Q/4πεR + C

I assume zero potential at infinity. If V(∞) = 0 , C = 0.

At point A, potential difference will be:
|r - r1| = |(0,0,0) - (1,0,0)| = |(-1,0,0)| = 1
V(1,0,0) = [1/ 4π(10^-9 /36π) ] *[q/1]

At point B, potential difference will be:
|r - r2| = |(0,0,0) - (-2,0,0)| = |(2,0,0)| = 2
V(-2,0,0) = [1/ 4π(10^-9 /36π) ] * [q/2]So for the difference in potential of point B relative to point A using the path integral, I just do V(-2,0,0) - V(1,0,0)?

Please tell me if I am doing the correct step. This is very important. Thank you

 

[ehild]

You get the work as the negative potential difference, but calculate it with the line integral this case. Integrate the electric field strength between the points along any line. (A straight line is the simplest).

What do you mean on [1/ 4π(10^-9 /36π) ]? Is it k=9.10^9? ehild

 

[nobrainer612]

yea , that is k. And why do we need to find W first? Can we just find V by using V=-∫E.dl ?

 

[ehild]

I meant work on unit positive charge: ∫E.dl=-ΔV

ehild

 

[paranormal]

.

I can provide you with a more comprehensive and accurate solution to this problem. First, let's define some terms and equations that are relevant to this problem:

- Electric potential (V): This is a scalar quantity that represents the electric potential energy per unit charge at a given point in space, measured in volts (V). It is defined as the amount of work needed to move a unit positive charge from infinity to that point, without any acceleration.

- Electric potential difference (ΔV): Also known as voltage, this is the difference in electric potential between two points in space, measured in volts (V). Mathematically, it is defined as the change in electric potential energy per unit charge between two points.

- Electric field (E): This is a vector quantity that represents the electric force per unit charge at a given point in space, measured in newtons per coulomb (N/C). It is defined as the force that a test charge would experience at that point.

- Coulomb's Law: This law states that the magnitude of the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as F = k*q1*q2/r^2, where k is the Coulomb constant (9x10^9 Nm^2/C^2), q1 and q2 are the two charges, and r is the distance between them.

- Path Integral: In this context, the path integral refers to the integral of the electric field (E) along a given path (dl). Mathematically, it can be expressed as ∫E°dl, where ° represents the dot product.

Now, let's move on to solving the problem. We are given a point charge (q) located at the origin, and we need to find the difference in potential between point B and point A using the path integral. To do this, we can follow these steps:

1. Calculate the electric potential at point A (Va) and point B (Vb) using the formula V = q/4πεr, where r is the distance between the point charge and the point in question. In this case, r for point A is 1 meter (since it is located at (1,0,0)) and r for point B is 2 meters (since it is located at (-2,0