# The capacitance of an isolated spherical conductor

I just started my topic on capacitance today. In the above derivation of capacitance of an isolated conducting sphere,+q charge is distributed uniformly. After substituting the value of potential, we get capacitance of the sphere.

Here is my question. If a -q charge is distributed uniformly the potential will be negative (V=-Kq/r). Now if I substitute this in the formula for capacitance I get a negative value but when I looked it up online it said that there is no negative value of capacitance. Where am I going wrong? What would be the capacitance of a negatively charged isolated conducting sphere?

• A (filled) sphere can't be conductive and uniformly charged at the same time. You are either talking about a (hollow) spherical shell, or the charge is uniformly distributed only on the surface (there will be no charge in the interior of the sphere).
– Puk
May 30, 2022 at 18:14
• @Puk A sphere is a hollow shell (in the standard mathematical terminology). If it is filled in it is a ball. Not that there is any problem in using the non-standard, but common, terminology that you use; I'm just saying that AJknight used correct terminology, and that it is clear that they meant $q$ is uniformly distributed on a spherical surface.
– ummg
May 30, 2022 at 18:31
• Just a small nitpick: voltage and capacitance are both scalar quantities. So your formulas should be, $C=4\pi\epsilon_0r$, etc. with the $r$ unbolded May 30, 2022 at 21:15
• It might help to consider the mechanical analogy. First write $V=\frac{1}{C}q$. Then compare with $F=-kx$. The spring constant is positive regardless of whether we stretch the spring or compress the spring. Aug 26, 2022 at 20:35

The sign of charge doesn't matter and the capacitance is positive. If the charge is $$-q$$, $$V = \frac{-q}{4\pi\epsilon_0 r}$$ $$C = -q/V = 4\pi \epsilon_0 r .$$

• So do you mean that if I am dealing with a negatively charged isolated conducting sphere I should keep the value of q as negative so that the two negatives cancel out and I get a positive value of capacitance? May 31, 2022 at 2:27
• @AJknight You can do that, but it doesn't matter whether you use a positive or negative charge because capacitance only depends on the geometry, not on the charge.
– Puk
May 31, 2022 at 2:57
• @Puk: Yes, the capacitance does depend on geometry and it can be calculated from charge and voltage without knowledge of the geometry. So the polarities are important. Aug 26, 2022 at 21:19

The polarity of the voltage is connected to the polarity of the charge. $$V = \frac{q}{4\pi\epsilon_0 r}$$

If the charge $$q = 10$$C and the voltage $$V = 5$$V, for example.

$$C = \frac{10C}{5V} = 4\pi\epsilon_0 r =2$$F

If the charge $$q = -10$$C, then voltage $$V = -5$$V.

$$C = \frac{-10C}{-5V} = 4\pi\epsilon_0 r =2F$$

• Note: $V<0$ says “V is negative”. However, $-V$ means “$(-1)V$” which is negative only when $V>0$. Aug 26, 2022 at 20:30
• @robphy: Ok. I will fix Aug 26, 2022 at 20:49