How does distance affect potential in an electric field?

[anmolnanda]

in which case potential is distance times the electric field..please explain so that i am cleared the the basics of this.

 

[silmaril89]

The potential is not the distance times the electric field. You probably think that based on the fact that distance times the electric field will give you the same dimensions as a potential. Also, you must be careful to say that this is an electrostatic potential, because the electric potential differs from the electrostatic potential. I'm going to assume you want to know how the electrostatic potential is related to the electric field. That is very simple,

<br /> \vec{E} = - \vec{\nabla}V<br />

 

[Doc Al]

in which case potential is distance times the electric field..please explain so that i am cleared the the basics of this.

For simple cases, where the field is constant, you can find the potential difference by multiplying the field times the distance:

ΔV = -Ed

Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c2"

 

[Nirgal]

<br /> \vec{E} = \vec{\nabla}V<br />

Correction: \vec{E} = -\vec{\nabla}V. Anyways, I'm sure you knew that.

anmolnanda it's important to understand the physical significance and the mathematical description of the electrostatic field vs the electrostatic potential field.

The Equation above says that the Electric Field is just the negative gradient of the electrostatic potential field. This means that the E field provides a description of how the Potential field changes in space. In particular, since the E Field is a gradient of the Potential then it points in the direction of maximum change (up hill or down hill) of the potential. Since the E field is actually the negative gradient then the E field points in the direction of maximum change down hill (towards the minimum of the potential).

(I'm just going to say E field and potential for now on knowing that I am referring to the electrostatic electric and potential fields).

You might ask why does the E field point down hill? Well the Potential field of a charge q is given by

V = kq/r where k is a constant, q is the charge and r is the radial distance from the charge. So V = V(r), that is, the potential is a function of position and there for defines a scalar field in space. So since it defines a scalar field in space and since it depends on the polarity of the charge (positive or negative) then there will be regions in space that have very negative potentials and regions with very positive potentials. So, if we draw lines pointing from the positive potential regions to the lower potential regions in such a way that the path of the lines are along the maximum change in potential, then you would have drawn the electric field lines from the positive charges to the negative charges (since the electric field is the change in the potential and so points along these lines).

So if you wanted to know what the so called "Voltage", or difference in Potential was between two different regions in space then you might need to know what the Potential field looked like. Say the potential field is given by V(r) and say that the points that we are interested in are V(A) and V(B) where A is a point of positive potential and B is a point of negative potential.

Then we could find out the difference in the potential between these two regions by writing down

V(A) - V(B) = Sum of the change in the potential in a tiny interval along the line of maximum change multiplied by the length of the tiny interval.

Now mathematically since we're talking about tiny intervals that brings us to differential calculus and since we're talking about summing up an enormous number of tiny intervals then that reminds us of integrals.

So

V(A) - V(B) = \int (Change in Potential)*(small-change-in-distance)

Ok but we said that the negative change in the Potential was the Electric field, let's call it E(r), and let's call the small change in distance dr. Then what we have is this:

V(A) - V(B) = \int -E(r)*dr

So \DeltaV = \int -E(r)*dr, where \DeltaV = V(A) - V(B).

Now consider that the Potential field changes in space in such a way that the Electric field is constant along lines of maxmimum change between two regions. In that case we can move the Electric Field outside the integral in the equation above and replace it with this,

\DeltaV = -E(r)\int dr = -Ed, where d = \int dr = distance between points A and B.

Therefore the difference in potential is just the product of the magnitude of the electric field multiplied by the distance between the two points multiplied by negative 1 to be consistent with signs.

Hope that helps.

 

[silmaril89]

Quite descriptive Nirgal, my bad on missing the minus sign.