# Electric potential -- magnitude of charges

#### logearav

Joined Aug 19, 2011
243
Revered members,
Electric potential is defined as the work done in moving a test charge from one point to other against the electric force. Usually the test charge is unit positive charge. What about the magnitude of the source charge?

#### BillO

Joined Nov 24, 2008
990
Revered members,
Electric potential is defined as the work done in moving a test charge from one point to other against the electric force. Usually the test charge is unit positive charge. What about the magnitude of the source charge?
Source charge? Do you mean the "charge" creating the force?

#### MrChips

Joined Oct 2, 2009
26,064
The magnitude is in electrons. For positive charge, it is the missing electrons.

#### logearav

Joined Aug 19, 2011
243
Source charge? Do you mean the "charge" creating the force?
Yes, Mr.BillO

#### russ_hensel

Joined Jan 11, 2009
825
The test charge should be imagined as approaching zero, else its position and charge will change the fields already present. In practice it is hard to work with a charge of 0. Units can be anything as long as they are consistent, the equations take into account the magnatude of the test charge. In nature we find charges only in multiples of the electron charge ( ignoring esoteric particles ).

#### steveb

Joined Jul 3, 2008
2,436
The test charge should be imagined as approaching zero, else its position and charge will change the fields already present.
This is indeed the traditional definition of a test particle, but the distinction turns out to not be critical from a practical point of view. While a large test particle does disturb the net field from the point of view of an external observer, it does not disturb the net field that effects it's own motion, or it's own net force (at least for a fixed field due to fixed charges). The reason is that the particle can not place a net force on itself. Hence, the field contribution from the particle does not come into play in a measurable way, as far as the test particle is concerned. The original field is still present and it is the only field that can affect the particle in terms of measurements an experimentalist might do to determine the original field value.

I guess the traditional definition is preferred in that it is conceptually simpler and one can think about putting more than one test charge in place, at one time. Also, the above also has the underlying assumptions that the original field is made from fixed charges that will not be induced to move by the test charge.

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#### BillO

Joined Nov 24, 2008
990
Yes, Mr.BillO
Then, as long as the force, or function describing the force, at the point of interest is given, the charge creating that force is of no importance. On the other hand, if the 'source' charge is fully described, one could calculate the resulting field and determine the force at the point of interest from that. However, in that case, the 'charge' is already given.

Generally, it would be difficult, or nearly impossible, to determine the source charge from only the function describing the force.

• logearav

#### logearav

Joined Aug 19, 2011
243
Now i have two points with charge +1C and +3C. Can this termed as potential difference? Because both are positive, i get this doubt.

#### steveb

Joined Jul 3, 2008
2,436
Now i have two points with charge +1C and +3C. Can this termed as potential difference? Because both are positive, i get this doubt.

As you said, electric potential is the work done (per unit charge) in moving a test particle between two points. Can you move a small positive test charge from the 1 Coulomb charge to the 3 Coulomb charge without doing any net work on the particle? Can you go the other way without net work being done on the particle?

Think about two hills, with one being taller that the other. Objects placed at the top of each hill have higher gravitational potential energy than similar objects placed in the valley between them. But, there is still a potential energy difference between those two peaks. I can get into a frictionless car at the top of the tallest hill and coast down into the valley and back up to the top of the smaller hill, and I will still have kinetic energy, which then requires me to put the breaks on to stay at the top of the small hill. If I then turn around and coast back down the smaller hill, I won't make it back to the top of the taller hill.

#### russ_hensel

Joined Jan 11, 2009
825
Then, as long as the force, or function describing the force, at the point of interest is given, the charge creating that force is of no importance. On the other hand, if the 'source' charge is fully described, one could calculate the resulting field and determine the force at the point of interest from that. However, in that case, the 'charge' is already given.

Generally, it would be difficult, or nearly impossible, to determine the source charge from only the function describing the force.
But in the case of statics you could construct the field and use maxwell's equation to fine the charge density ( divergence of field I think ).