# Potential difference: positive or negative?

1. Oct 5, 2015

### gracy

Why potential at a point can only be obtained by supposing/assuming a positive test charge there?
For example
There was a question
In figure two points A and B are located in a region of electric field.,The potential difference $VB$-$VA$ is
1-positive
2-Negative
3-zero
4-none of the above
Apparently the source charge is positive

we will only get the answer to be 2-Negative when we would assume positive test charge but if we assume negative test charge at A &B we are going to get answer 1-positive.So is it like we are bound to assume positive charge at points mentioned in question to get whether potential difference is positive or negative?

2. Oct 5, 2015

### Staff: Mentor

Yes. The "test charge" is always positive, by definition.

3. Oct 5, 2015

### ehild

The figure shows the electric field lines.
How is the electric field at a point defined?

4. Oct 5, 2015

### gracy

Force per unit charge.
The electric field at a given point is defined as the (vectorial) force that would be exerted on a stationary test particle of unit charge by electromagnetic forces

5. Oct 5, 2015

### ehild

And what is that test particle?What does unit charge mean?
http://www.thefreedictionary.com/electric+field+strength

Last edited by a moderator: Apr 17, 2017
6. Oct 6, 2015

### gracy

And we are told to put negative charge there(in electric field ) in that case we can ,right?

7. Oct 6, 2015

### ehild

You can put a negative charge into any electric field. Do it. And what do you do after?

8. Oct 6, 2015

### gracy

But it can not be called "test charge",right?

9. Oct 6, 2015

### Staff: Mentor

Correct. A test charge is always positive.

10. Oct 8, 2015

### Chandra Prayaga

The potential difference between points A and B (or any two points) does not depend on the test charge used. The relation is standard

V_B - V_A = - \int_A^B \overrightarrow{E}.d\overrightarrow{s}

You can use a negative or positive test charge and multiply both sides of the above relation with the charge. The equation will change to a relation between the difference of potential energies, and the work done by the electrostatic force. In case my Latex is not clear, there is a dot product between the electric field and the infinitesimal displacement in the integral, and the integral is over any path joining points A and B. It does not matter whether you "call" it a test charge or not. The fact is that you can use either positive or negative charges.