# Definition of Voltage using integral: minus sign?

So I'm studying electrostatics and I came across to two different definitions of potential difference/voltage (because we're in stationary regimes) and I'm having trouble understanding how the expressions are equivalent.

They are for a voltage between point A and point B

$$U=V_a - V_b =\int_{a}^{b} \textbf{E} \cdot d\textbf{s}$$

and, on the other hand,

$$U= V_b - V_a = - \int_{a}^{b} \textbf{E} \cdot d\textbf{s}$$

How can both of this expressions represent the potential difference between points A and B? Aren't they symmetric?

• Came across where? – Qmechanic Sep 22 '18 at 18:48

## 2 Answers

From the relation $$\mathbf{E} = - \nabla V$$ and the gradient theorem, we can see that $$\Delta V = \int_a^b \mathbf{E} \cdot d \mathbf{s} = \int_a^b (-\nabla V) \cdot d \mathbf{s} = \int_b^a (\nabla V) \cdot d \mathbf{s} = V(a) - V(b)$$ So the first expression is mathematically correct. The electric field points from higher potential to lower potential. Let $$a$$ be at a higher potential than $$b$$. If you integrate the electric field from $$a$$ to $$b$$, it points in the same direction as the path, and the dot product will thus give a positive answer, in agreement with $$V(a) - V(b)$$. The second one represents the potential difference between $$a$$ and $$b$$, which represents the work required to move a charge form $$a$$ to $$b$$, which is opposite to the field, which gives $$V(b) - V(a)$$. So it is better to stick with the second one.

• I disagree. I would understand "voltage between point A and point B" to mean $V(b)-V(a)$ i.e. the potential difference in going from A to B. – Emilio Pisanty Sep 22 '18 at 17:46
• @Emilio Pisanty: No, $V(b) - V(a)$ is the work required to move a charge form $a$ to $b$, which is opposite to the field. However I agree that the potential difference should be the negative of $V(a) - V(b)$, I have changed my answer – user7777777 Sep 22 '18 at 17:59

The first equation can be written as $$V_a - V_b = \int_a^b {\bf E}\cdot d{\bf s} =-\int_b^a{\bf E}\cdot d{\bf s}\,.$$ This follows from a rule in calculus. It is the potential difference $$V_a -V_b$$, i.e. it is the work done per unit test charge by an external force in moving the test charge from point $$b$$ to point $$a$$ with no change in the kinetic energy of the test charge.

The second expression $$V_b - V_a = - \int_a^b {\bf E}\cdot d{\bf s}$$

is the potential difference $$V_b - V_a$$ and it is the work done per unit test charge by an external force in moving the test charge from point $$a$$ to point $$b$$ with no change in the kinetic energy of the test charge.

So the two expressions are related but the author of the text you are reading from seems to have made an error and was not clear in explaining the meanings of the two expressions. I also apologise for my earlier incorrect answer.

• In what way is $\int_a^b {\bf E}\cdot d{\bf s} = -\int_a^b{\bf E}\cdot d{\bf s}$, as the two expressions in the question suggest? The two expressions can't be equal as they differ by a negative sign. Switching $a$ and $b$ does not change this fact in any way. – user7777777 Sep 22 '18 at 17:25
• I am in the processing of editing my original answer. I think I made a mistake. Sorry. – Physics_Et_Al Sep 22 '18 at 17:31