# Explanation of the negative integral

As referred to this question Why can't charge be in a stable equilibrium in electrostatic field?, it is given that

Fig. 5–1.If $P_0$ were a position of stable equilibrium for a positive charge, the electric field everywhere in the neighbourhood would point toward $P_0$. Consider a tiny imaginary surface that encloses $P_0$, as in Fig. 5–1. If the electric field everywhere in the vicinity is pointed toward $P_0$, the surface integral of the normal component is certainly not zero. For the case shown in the figure, the flux through the surface must be a negative number. But Gauss’ law says that the flux of electric field through any surface is proportional to the total charge inside. If there is no charge at $P_0$, the field we have imagined violates Gauss’ law. It is impossible to balance a positive charge in empty space—at a point where there is not some negative charge. A positive charge can be in equilibrium if it is in the middle of a distributed negative charge. Of course, the negative charge distribution would have to be held in place by other than electrical forces.

Please explain the part which is denoted in bold. I have tried to comprehend the idea but did not get anything. My question is why the integral will be negative. I was also reading about this in the book Basic Laws of Electromagnetism, I encountered the same problem where is written:

the flux of E through the surface S is negative

Any help will be appreciated.

• You want to revisit the meaning of a surface integral. This might help. Apr 18 '18 at 8:17
• @theboywhobelieved, it did not help me to understand my problem, it will be very helpful if you give me some clue to understanding the matter. Apr 18 '18 at 8:29

Consider the following example: You have an arbitrary surface $S$ and a quantity $\vec{F}$ that points through it. You can calculate the flux $\Phi$ by integrating the quantity $\vec{F}$ over the area of the surface $S$. The quantity has a direction and the surface is oriented. You end up with: $\Phi_1 = \int_S \vec{F}\cdot d\vec{A}$
If you now reverse the direction of the quantity, $\vec{F}$ becomes $-\vec{F}$ and you will obtain a minus sign in your flux: $\Phi_2 = -\int_S \vec{F}\cdot d\vec{A}$.
Do you agree that if the flux is non-zero, either $\Phi_1$ or $\Phi_2$ have to be negative? 