# Potential Difference

1. Jan 25, 2008

### breez

By convention, potential difference is defined as the $$\frac{W}{q_0}$$, where $$q_0 > 0$$ correct? When computing potential differences in my textbook, states in the derivation of the relationship that $$V_f - V_i = $\int_i^f E\dot ds$$$ by assuming a positive charge. From the relationships of potential and charge, seems if we used a negative charge, we would compute negative values of difference of potential in comparison to using a positive charge.

**that's a dot product in the integral. I don't know the LaTeX for a dot.

2. Jan 25, 2008

### PiratePhysicist

I'm not sure what you're asking. Can you state it more explicitly?

3. Jan 25, 2008

### breez

Basically my question is just the first sentence. When we say difference in potential between 2 points in an electric field, we are assuming by convention that the test charge is positive correct? And using a negative charge would yield the opposite signed potential difference in respect to the case where a positive charge is used.

More explicitly, $$V_f - V_i = \frac{W}{q_0} = -$\int_i^f E\dot ds$$$ is the equation used to calculate the potential difference between points i and f in an E-field. By convention we assume the charge q_0 is positive correct?

4. Jan 25, 2008

### PiratePhysicist

Where is $$\frac{W}{q_0}$$ coming from?
The definition of W is usually work, which is the amount of energy required to move from one position to another, which in this case would be the difference of potential energies of two positions around the charge so: $$W=V_f - V_i$$ like you had in your first post. This is a definition that is applied in many other areas besides just electromagnetism, so there is no charge involved here. What charge will do is change which is larger the final potential or the initial potential.

5. Jan 25, 2008

### Rainbow Child

The potential energy $U$, is defined by $$W_{elect}^{i\rightarrow f}=U_i-U_f$$

The potential $V$ and the difference, are defined by

$$V_A=\frac{U_A}{q}\Rightarrow V_i-V_f=\frac{W_{elect}^{i\rightarrow f}}{q}$$

no matter the sign of the charge.

6. Jan 25, 2008

### breez

For example when we say that the potential difference between point f and point i is 50 V, then 50 V with respect to a positive charge moving from point f to point i. The potential difference would be -50 V if it were a negative charge moving from f to i. I'm asking if it's standard to assume a positive charge when we talk about potentials differences.

Also, rainbow child, I believe in your last formula you are missing a negative sign in front of the fraction.

7. Jan 25, 2008

### Hurkyl

Staff Emeritus
\cdot (center dot) ($\cdot$). Use \cdots to get multiple dots, ellipsis-style; also \ldots (lower dots) ($\ldots$). The command you issued, \dot, acts as an accent; it places a dot over the next symbol, as you have observed.

8. Jan 25, 2008

### Rainbow Child

The potential does not depend from the charge that moves in the electrical field. The work done on the charge depends from the charge.
Thus if the potential difference from point A to point B is $$V_{AB}=50V$$ then the work on a charge q=+1 Cb is W=50 J, and on a charge q=-1 Cb is W=-50 J.