0
$\begingroup$

For an spherical conductor shell of radius R, it is known that

$V (r)= \dfrac {q}{4 \pi \epsilon_o r}$ for r>R

$V(r) = \dfrac {q}{4 \pi \epsilon_o R}$ for r<=R

My texbook gives the capacitance as

$C = \dfrac {q}{V}= 4 \pi \epsilon_o R$

Why is the V in the surface been used for it? If I used, the expresion for r>R , lets say r=2R, And calling V the potencial at the surface(r=R) and V' the potential at r=2R then

$V'= V (r=2R)= \dfrac {q}{4 \pi \epsilon_o 2R}= \dfrac {1}{2}(\dfrac {q}{4 \pi \epsilon_o R})=\dfrac {V}{2}$

then since the charge is constant but the potential is different, the capacitance would be: $ C'= \dfrac{q}{ V'}=2 \dfrac{q}{ V}=2C$, which makes the capacitance dependent on the potential... Why is this wrong? How do I make sense of the capacitance of an isolated object, I thought one always needes a pair of objects placed one close to the other to have a capacitance. The book does not specify if its a solid sphere or an spherical surface(as I am assuming). In case it is a solid sphere the inner potential would be quadratic, would the result be different ?

$\endgroup$
1
$\begingroup$

The capacitance measures the charged stored ($Q$) per unit voltage between two oppositely-charged conductors (carrying $Q$ and $-Q$). The charge of an isolated conductor is a special case of this, where the second conductor is taken to be located an infinite distance away. (You can imagine the second conductor being a sphere of radius $r\rightarrow\infty$, although the shape of the distant conductor does not actually matter.)

So for an isolated conductor, the inverse capacitance ($C^{-1}$) is the voltage difference between the conductor and infinity, divided by $Q$. The voltage at infinity is zero by convention*, and the voltage at the other conductor is the voltage of the sphere in your case, $V(R)=Q/(4\pi\epsilon_{0}R)$. So the relevant voltage difference is just $V(R)$ and $C=4\pi\epsilon_{0}R$.

*Even when there is taken to a total charge $-Q$ at infinity, the potential at $r\rightarrow\infty$ is still vanishing, because the charge is spread out over a sphere with infinite area. You can see this by taking the outer sphere to be at a finite radius $b$; then the voltage difference between the conductors at $R$ and $b$ is $Q/(4\pi\epsilon_{0}R)-Q/(4\pi\epsilon_{0}b)$ which goes to just $V(R)$ as $b\rightarrow\infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.