Potential difference: positive or negative?

[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 ##V_B$-$V_A## is
1-positive
2-Negative
3-zero
4-none of the above
the answer is 2-Negative
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?

 

[Dale]

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

 

[ehild]

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

 

[gracy]

How is the electric field at a point defined?

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

 

[ehild]

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

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

 

[gracy]

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

 

[ehild]

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

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

 

[gracy]

You can put a negative charge into any electric field.

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

 

[Dale]

Correct. A test charge is always positive.

 

[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.