# Homework Help: Understanding the potential

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1. Oct 11, 2016

### diredragon

1. The problem statement, all variables and given/known data
I dont exactly conceptually understand what the potential is. The electric field is a vector at a given point from the charge that describes the strenght and the direction of the fields. Potential is given as a potential of a point in referenve to another point in spave? I have uploaded the picture od a segment of my lecture today where it discribes that there exists a constant in the potential formula which dissapears when we introduce the potential difference..
2. Relevant equations
3. The attempt at a solution

Can someone give a little insight in what it is if you understand what i mean? The potential of a point A is known but the potential of A' is potential of point A plus some constant which is electric field line integral from R to R'. How is that constant if R' is arbitrarily chosen?

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2. Oct 11, 2016

### BvU

Hi there,

Are you familiar with the gravitational potential ?

3. Oct 11, 2016

### diredragon

Kinda.. it is property of a gravitational field at some point in space..right? I know the expression but dont get the concept..

Last edited: Oct 11, 2016
4. Oct 11, 2016

### BvU

I meant the simpler version: potential energy = mgh. The difference in potential energy only depends on the difference in height, not on the path taken to get from the one height to the other.

Basically a conservative force is the derivative of a potential.

the potential of A' is potential of point A plus some value which is electric field line integral from R (position of A) to R' (position of A').

So R' is not arbitrarily chosen but the path from A to A' can be arbitrarily chosen.

5. Oct 11, 2016

### diredragon

So lets say we have a charge $q$ and a point $A$ in space whose potential is $V_A=\frac{Q}{4πε_or_A}$ where $r_A$ is the vector from $Q$ to $A$ .If the above is also true for $B$ but with $r_B$ then the potenital defference between these two is some value $V_{B-A}=\frac{q}{4πε_or_{B-A}}$ What good is this expression? What use do we have of this? The potential difference between those two points..?

6. Oct 11, 2016

### BvU

If A has $V_A=\frac{Q}{4πε_or_A}$ and B has $V_B=\frac{Q}{4πε_or_B}$ then the potential difference between these two is of course $$V_B - V_A = \frac{Q}{4πε_or_B} - \frac{Q}{4πε_or_A}$$
Which is the energy per charge needed to bring a charge from A to B