# How to derive formula for capacitance of spherical capacitor

1. Jan 4, 2016

### Dexter Neutron

Please explain in detail how to derive formula for capacitance of spherical capacitor?

2. Jan 4, 2016

### ehild

A single metal sphere of radius R having charge Q has the potential U=kQ/R with respect to infinity, so its capacitance is C = Q/U = R/k.

3. Jan 5, 2016

### Staff: Mentor

A Google search for "capacitance of spherical capacitor" gave me this as the first hit:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html

Note that unlike ehild's example, this assumes the capacitor is a conducting sphere (one "plate") surrounded by a conducting spherical shell (the other "plate"). You should specify which version of "spherical capacitor" you meant.

In either case, it would be best for you to find a derivation for the situation that you want, and try to understand it, and then ask specific questions about parts that you don't understand; instead of asking us to give you a complete derivation which would be a lot more work for us, and is unlikely to be better than what you can find in a textbook or with a Google search.

4. Jan 5, 2016

### Dexter Neutron

In my textbook it is given that
Consider a small sphere of radius r2 having -ve charge of magnitude q enclosed by a large sphere of radius r1 having a +ve charge with magnitude q. Assume an imaginary sphere at a distance r between both the spheres. The flux through sphere = q/ε0
E x A = q/ε0
Thus E = q/Aε0 = q/4πr2ε0
Now they integrated this electric field and found out potential difference between both the spheres.
What my doubt is that Why they only considered the electric field due to negatively charged sphere and did not consider the field due to positive charge.
The positive charge must also affect the electric field at that point so it must also be considered.
Now my 2nd doubt:
Potential difference between two points is the amount of energy per coulomb required to take a test charge from one point to other.
Since we can simply find potential at two different points and subtract them to get the potential difference then while doing that here
Potential at smaller sphere = potential due to charge on smaller sphere + potential due to poitive charge on larger sphere
= -kq/0 + kq/r1-r2
this goes into indeterminate form.Is there any other method other than this and that given in textbook to arrive at the result properly.
Thanks.

5. Jan 5, 2016

### nasu

Did you study Gauss' law yet?

If not, what do you know about the electric field inside a conductor or conducting shell?

6. Jan 6, 2016

### Dexter Neutron

I have studied guass law but I want to ask that the electric field obtained from guass law is just due to the negative charge and not due to, positive charge.We must also take electric field due to positive charge as it must also contribute towards potential difference then why they did not consider it?

7. Jan 6, 2016

### vanhees71

The most simple approach is to use the local laws. For electrostatics this means to evaluate the electric potential, obeying the equation
$$\Delta V=-\frac{\rho}{\epsilon}.$$
For your example of a spherical capacitor you introduce spherical coordinates. Due to symmetry you can assume that $V=\Phi(r)$, i.e., it depends only on the distance from the origin, which is the center of the concentric spheres. Then between the plates there are no charges. So you have
$$\frac{1}{r} (r V)''=0.$$
Now by successive integrations you get
$$(r V)'=C_1 \; \Rightarrow \; r V =C_1 r+C_2 \; \Rightarrow \; V=C_1+\frac{C_2}{r}.$$
Now let the radius of the inner sphere be $a$ and that of the outer $b$. The voltage difference may be $U$. Then you can set $V(a)=0$ and $V(b)=U$. That means you have
$$C_1+\frac{C_2}{a}=0, \quad C_1+\frac{C_2}{b}=U \; \Rightarrow \; C_1=-\frac{C_2}{a} \; \Rightarrow \; C_2=-U \frac{ab}{b-a}.$$
Since we assume that the capacitor carries opposite charges of the same magnitude on both spheres, together we have
$$V(r)=\frac{Ub}{b-a} \left (1-\frac{a}{r} \right ).$$
The electric field is directed radially out with the radial component given by
$$E_r(r)=-V'(r)=-\frac{ab U}{(b-a)r^2}.$$
To get the charge on the outer shell, integrate the electric field over a spherical shell enclosing it. Only the inner shell with radius $R$, with $R \in (a,b)$ arbitrary, gives something different from 0, and thus
$$Q=-E_r(R) 4 \pi R^2=\frac{4 \pi \epsilon ab}{b-a} U\; \Rightarrow \; C=\frac{Q}{U}=\frac{4 \pi \epsilon ab}{b-a}.$$

Last edited: Jan 6, 2016
8. Jan 6, 2016

### Dexter Neutron

Thanks for your help but it is too difficult to understand.Could you please explain me first 3 steps?

9. Jan 6, 2016

### Staff: Mentor

So you know (or should know) that Gauss's law in its integral form uses the total charge inside an imaginary closed "Gaussian surface".
For your Gaussian surface, imagine a sphere whose radius is between r1 and r2. How much of the positive charge is enclosed by that Gaussian surface?

10. Jan 6, 2016

### vanhees71

Do you have vector calculus and the differential operators div, grad, rot and the Laplace operator. If not, my answer is perhaps useless. Then you must use Gauss's Law in integral form as suggested by jtbell in the previous posting.

11. Jan 6, 2016

### Dexter Neutron

I am getting confused.Please tell me actually what is the potential difference here i.e. between which two points are we taking potential difference.

12. Jan 6, 2016

### nasu

Each of the two spheres have the same potential everywhere on its surface.
So the difference is between the outer surface (you can take any point on it) and the inner ball (any point on the inner ball).
For simplicity you can imagine the two points on a radial line, of course.

The field inside a charged, metallic shell is zero, at equilibrium (electrostatic conditions). Also zero inside a metallic ball.
You should get familiar with these basic concepts. You can "prove" them as basic applications of Gauss' law (integral form).