Reason for adding the negative sign in the expression of electrostatic potential energy for the differential element of distance

Question

My physics professor taught us that whenever we express the electrostatic potential energy, we usually write the expression as,

$$\displaystyle\int_\infty^0\dfrac{Q}{4\pi\epsilon_0r^2}\,(-dr)\cos0.$$

He explained that the differential element of distance here is dr and because it being 'by nature' negative, the negative sign is multiplied so that we keep it positive.

The phrase, 'by nature' at least according to him means that the differential element is negative in sign as a matter of fact.

His explanation seems to be vague, and I do not get an intuitive understanding of

Why is the negative sign used?

Now I already know the expression,

$\displaystyle \int_{position_a}^{position_b}(-q\vec{E}).\vec{dl}$

and why we add the negative sign here, but other than the same the expression with $-dr$ confuses me

Answer 1

The negative sign is added to show that work done by the electric field is negative when you move a charge against the field. As shown in the definition $U = -\int \vec{F} \cdot d\vec{r}$ which you can find a great explanation on in the Feynman Lectures on Physics. Volume 2, chapter 4.

Answer 2

In electrostatics, the electric field $\vec{E}(\vec{x})$ satisfies the differential equation $\vec{\nabla} \times \vec{E}(\vec{x})=0$ for all $\vec{x} \in \mathbb{R}^3$. As a purely mathematical fact, this is equivalent to the statement that the vector field $\vec{E}(\vec{x})$ can be be written as the gradient of some scalar function, i.e. $$ \vec{E}(\vec{x})=- \vec{\nabla} \phi(\vec{x}), \tag{1} \label{1} $$ where $\phi(\vec{x})$ is referred to as the electrostatic (scalar) potential. The choice of the minus sign in the definition of the potential in eq. \eqref{1} is, of course, a mere convention (maybe the physisists of some intelligent civilization in the Andromeda nebula have chosen a plus sign instead). For a given electric field $\vec{E}(\vec{x})$, the associated potential $\phi(\vec{x})$ is only defined up to an arbitrary additive constant: the potentials $\phi(\vec{x})$ and $\phi^\prime(\vec{x})=\phi(\vec{x})+c$ describe the same electric field.

How can one find the scalar potential $\phi(\vec{x})$ associated with some given electric field $\vec{E}(\vec{x})$? This is achieved by computing the line integral $$ \int\limits_{\vec{x}_0}^\vec{x} \! d \vec{x}^\prime \cdot \vec{E}(\vec{x}^\prime)= - \int\limits_{\vec{x}_0}^\vec{x}\! d \vec{x}^\prime \cdot \vec{\nabla} \phi(\vec{x}^\prime)=- \phi(\vec{x})+\phi(\vec{x}_0), \tag{2} \label{2}$$ where $\vec{x}_0$ is an arbitrarily chosen point. Note that the line integral on the left-hand-side of eq. \eqref{2} is independent of the curve connecting the intial point $\vec{x}_0$ and the endpoint $\vec{x}$. We have thus found the formula $$ \phi(\vec{x})= -\int\limits_{\vec{x}_0}^\vec{x} \! d \vec{x}^\prime \cdot \vec{E}(\vec{x}^\prime) +\phi(\vec{x}_0) \tag{3} \label{3} $$ As an example, let us consider the Coulomb field $$ \vec{E}(\vec{x}) = \frac{Q}{4 \pi \epsilon_0} \frac{\vec{x}}{|\vec{x}|^3} \tag{4} \label{4}$$ of a point charge $Q$ sitting at the origin. We choose the path $\vec{x}^\prime(r)= r \vec{x}/|\vec{x}|$ between $\vec{x}_0 = \lim\limits_{r\to \infty} \vec{x}^\prime(r)$ and $\vec{x}= \vec{x}^\prime(|\vec{x}|)$. Using $$ d \vec{x}^\prime(r)= \frac{d \vec{x}^\prime(r)}{dr} dr= \frac{\vec{x}}{|\vec{x}|} dr \tag{5} \label{5}$$ we find \begin{align} \phi(\vec{x})&=- \int\limits_\infty^{|\vec{x}|} \! dr \, \frac{\vec{x}}{|\vec{x}|} \cdot \vec{E}(\vec{x}^\prime(r))+\phi(\infty) \\[5pt] &=- \frac{Q}{4\pi \epsilon_0} \int\limits_\infty^{|\vec{x}|} \! dr \, \underbrace{\frac{\vec{x}}{|\vec{x}|} \!\cdot \!\frac{\vec{x}}{|\vec{x}|}}_{1} \, \frac{1}{r^2}+\phi(\infty) \\[5pt] &=\frac{Q}{4\pi \epsilon_0 |\vec{x}|}+ \phi(\infty). \tag{6} \label{6} \end{align} With the usual choice $\phi(\infty)=0$, we have found the Coulomb potential $$ \phi(\vec{x})= \frac{Q}{4\pi \epsilon_0 |\vec{x}|}. \tag{7} \label{7}$$ We may check our result by computing $$ - \vec{\nabla} \phi(\vec{x}) = - \frac{Q}{4\pi \epsilon_0} \vec{\nabla} \frac{1}{|\vec{x}|}= -\frac{Q}{4\pi \epsilon_0 } \left(- \frac{1}{|\vec{x}|^2} \vec{\nabla} |\vec{x}| \right)=\frac{Q}{4\pi \epsilon_0 |\vec{x}|^2} \frac{x}{|\vec{x}|}, \tag{8}$$ recovering indeed \eqref{4}.

Let me add three final remarks:

1. As it stands, your first integral $$ -\frac{Q}{4\pi \epsilon_0} \int\limits_\infty^0 \! dr \, \frac{1}{r^2} \tag{9} $$ does not make sense. Because of the singularity of the integrand at $r=0$, the integral diverges (does not exist). However, with the replacement $0 \to |\vec{x}|$, you would recover the expression for $\phi(\vec{x})-\phi(\infty)$ as shown in the second line of eq. \eqref{6}.

2. Whenever you compute a line integral $$ \int\limits_C \! d\vec{x} \cdot \vec{F}(\vec{x}), \tag{10} \label{10} $$ you choose an arbitrary (differentiable) parametrization $t \to \vec{x}(t)$ with $a \le t \le b$ of your curve $C$. The integral \eqref{10} is defined by $$ \int\limits_C \! d\vec{x} \cdot \vec{F}(\vec{x}):= \int\limits_a^b \! dt \, \frac{d \vec{x}(t)}{dt} \cdot \vec{F}(\vec{x}(t)), \tag{11} \label{11}$$ being simply a one-dimensional integral with integration variable $t$. Note that \eqref{11} is, of course, independent of the choice of the parametrization of the curve $C$.

3. If you do not like the order of your limits of integration, you may always use the formula $$ \int\limits_a^b \! dt \, f(t) = -\int\limits_b^a \! dt \, f(t). \tag{12} $$

Answer 3

Electric potential energy can be equivalently understood as following:moving a charge from a certain point to the reference point(the infinity in your case), the work done by the electrical field. In your formula: $$ \int_{\infty}^{0} \frac{Q}{4\pi \epsilon_0 r^2} (-\mathrm{d}r) \cos 0 $$ The origin of coordinates is placed at the place where the charge of field $Q$ source stays. So $dr>0$ if you move the test charge away from $Q$, and $dr<0$ if you do the opposite, which is what the formula says. By the way, the concept of electrical potential energy is not that important as we usually concern about its variation!