# My conundrum with Gauss’ law in electrostatics

If I use Gauss’ law to calculate the electric field outside of a charged (conducting or insulating) sphere or a point charge, the fields are the same. However, as a test approaches a point charge, the field becomes infinite whereas the test charge approaching the surface of the sphere reaches a finite value of $Q/4πϵ_0R^2$, where $R$ and $Q$ are the radius and charge of the sphere. Why wouldn’t the field also go to infinity as I get infinitesimally close to the surface of sphere? Those are just “point charges” on the surface. What ultimately led me to this question was trying to physical understand why Gauss’ law predicts the same field for a charge point and charged sphere (obviously $r > R$). I don’t understand it.

• To see why the field is a constant on a sphere, imagine what happens as you coming really close to the surface of the conducting sphere? This value is defined in many textbooks. Oct 8, 2013 at 2:35
• Can you be more explicit. That's why I am asking the question. Oct 8, 2013 at 2:36
• Okay, I will tried and write it up. Oct 8, 2013 at 2:37

A physical way to see the answer is to ask what the test charge sees as it approaches the surface of the conductor. At first, with a test charge far from the charged sphere, it sees the sphere as point-like with a field value of $Q/4πϵ_0r^2$, where r is the radius of the Gaussian spherical shell. However, as the test charge gets really close to the surface of a conducting sphere, the surface of the sphere appears as an infinite plane whose field value is

$$E_{sphere}(r→R) = \frac{Q}{4πϵ_0R^2}=\frac{Q/4πR^2}{ϵ_0}=\frac{σ}{ϵ_0}$$

In other words, as the test charge approaches the conductor’s surface, it’s “field-of-view” is reduced and “sees” less field lines. This compensates for the diminishing influence of any particular piece, giving the electric field a fixed and constant value at the surface. In other words, at really small distances from the surface, the conductor is an infinite plane, as shown above.

• Nice answer. I like how you brought in the infinite plane. Oct 8, 2013 at 2:58

Those are just “point charges” on the surface.

Yes, except that the "point charge" you approach has an infinitesimal charge (thus cancelling out and giving a net zero field). There are infinite such points, true, but you don't approach all of them -- just one. The rest combine to give a finite value, $\frac{Q}{4\pi\epsilon_0R^2}$

Gauss law predicts the same field because the net enclosed charge is the same and has the same symmetries for $r>R$.

• How does the rest of the charges combine to give the finite value of the sphere's surface? Oct 10, 2013 at 3:02
• @LisaLee We do not approach the rest of the charges, so the $r$ for them is not zero. Each one is a finite distance away, and as each one is infinitesimal charge the force by each one is infinitesimal. In this case, the infinite sum of infinitesimals converges. Oct 10, 2013 at 4:17