# Electric potential due to multiple charges

1. Apr 22, 2014

### hasankamal007

Hello!

For Single Charge:
I studied the formula for electric potential for a point charge (V=kq/r). It was derived when the test charge approached the source charge from infinity "head-on". In this case the electric field due to source charge and displacement of test charge were vectors in same(or at 180degrees) direction. Fine.

For Multiple Charge:
Now, suppose there's a cube with +q point charges at its vertices. Now, to calculate the potential, let's say at the centre of the cube we add potentials due to all those eight charges using v=kq/r where we put r as the distance between vertice and centre of cube.

But this should work only if the test charge brought from infinity is head-on to every charge at vertice of cube! And you cannot do that since in this arrangement, if you try to bring the test charge from infinity head on to any one charge out of eight, you will start having an angle between other six charges.

Hence, we must take account of cos theta and stuff because then electric field (of source) and displacement (of test charge) are at an angle to each other. So why do we straightaway use that formula?

2. Apr 22, 2014

### Staff: Mentor

For any one of the charges, the electric field is radially (spherically) symmetric. So any non-radial components of differential displacement, when dotted with the electric field vector are zero. So only the radial component of differential displacement contributes to the work required to move a test charge from infinity to any arbitrary location. This applies to each and every charge of the multicharge array.

Chet

3. Apr 22, 2014

### hasankamal007

Exactly Chet!,
the dot product should zero out the perpendicular component of electric field on the test charge(W=F.s.cosθ).
So, we shouldn't use V=kq/r for a such-arranged multi charged system because this equation was itself derived only for head-on cases.
There should be something like cosθ involved somewhere due to the dot product. But still, only (kq/r+kq/r+....) is used everywhere regardless of this problem in such-arranged(cubic) multi charge system. So, what to do?

For instance, I was solving this question in my physics assignment I described earlier(charges on cube's vertex) AND you get correct answer if you use V=kq/r+kq/r...=8kq/r.
But now we're saying V=kq/r isn't valid here. Then, who's right?

4. Apr 22, 2014

### Staff: Mentor

The combined electric field from the several charges is equal to the linear superposition of the electric fields from the individual charges. So all we really need to be able to prove is that the work to bring a test charge from infinity to any arbitrary location relative to a single charge q is independent of the path we take. Now, if we put the single charge q at the origin, the electric field is:
$$\vec{E}=\frac{kq}{r^2}\vec{i}_r$$
Let $\vec{ds}$ be an arbitrary displacement of the test charge. Then,
$$\vec{ds}=(ds)_r\vec{i}_r+(ds)_θ\vec{i}_θ+(ds)_{\phi}\vec{i}_{\phi}$$
Now, in spherical coordinates, (ds)r = dr, so:
$$\vec{ds}=\vec{i}_rdr+(ds)_θ\vec{i}_θ+(ds)_{\phi}\vec{i}_{\phi}$$
If we take the dot product of this arbitrary displacement vector with the electric field vector, we get:
$$(\vec{E}\centerdot \vec{ds})=\frac{kq}{r^2}dr$$
Note that there are no contributions from the components of the displacements in the other two coordinate directions. So, the work to bring a test charge from infinity to any arbitrary location relative to the charge is independent of path.

Chet