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Dexter Neutron
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Please explain in detail how to derive formula for capacitance of spherical capacitor?
Start with the concept of capacitance: stored charge over potential difference.Dexter Neutron said:Please explain in detail how to derive formula for capacitance of spherical capacitor?
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?nasu said:Did you study Gauss' law yet?
If not, what do you know about the electric field inside a conductor or conducting shell?
Thanks for your help but it is too difficult to understand.Could you please explain me first 3 steps?vanhees71 said: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
$$Q=-E_r(R) 4 \pi R^2=\frac{4 \pi \epsilon ab}{b-a} U\; \Rightarrow \; C=\frac{Q}{U}=\frac{4 \pi \epsilon ab}{b-a}.$$
So you know (or should know) that Gauss's law in its integral form uses the total charge inside an imaginary closed "Gaussian surface".Dexter Neutron said:I have studied guass law
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?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?
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.Dexter Neutron said:Thanks for your help but it is too difficult to understand.Could you please explain me first 3 steps?
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.jtbell said: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?
A spherical capacitor is a type of capacitor where the two conducting surfaces are spherical in shape. It consists of two concentric spherical conductors separated by a dielectric material.
The formula for the capacitance of a spherical capacitor is C = 4πε0εrr, where ε0 is the permittivity of free space, εr is the relative permittivity of the dielectric material, and r is the radius of the spherical conductors.
The capacitance of a spherical capacitor is inversely proportional to the distance between the spherical conductors. This means that as the distance between the conductors increases, the capacitance decreases, and vice versa.
Yes, the formula is valid for any dielectric material, as long as the relative permittivity (εr) of the material is known.
The capacitance of a spherical capacitor is directly proportional to the size of the spherical conductors. This means that as the radius of the conductors increases, the capacitance also increases.