(a) Two spheres have radii a and b, and their centers are a...

(a) Two spheres have radii a and b, and their centers are a distance d apart. Show that the capacitance of this system is $$ C=\frac{4 \pi \epsilon_{0}}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}} $$ provided $d$ is large compared with $a$ and $b .$ Suggestion: Because the spheres are far apart, assume the potential of each equals the sum of the potentials due to each sphere. (b) Show that as $d$ approaches infinity, the above result reduces to that of two spherical capacitors in series.

(a) Two spheres have radii a and b, and their centers are a distance d apart. Show that the capacitance of this system is
$$
C=\frac{4 \pi \epsilon_{0}}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}}
$$
provided $d$ is large compared with $a$ and $b .$ Suggestion: Because the spheres are far apart, assume the potential of each equals the sum of the potentials due to each sphere.
(b) Show that as $d$ approaches infinity, the above result reduces to that of two spherical capacitors in series.

Key Concepts

Approximation Techniques in Electrostatics

In many practical problems, approximations such as assuming that the distance between objects is much larger than their sizes are used to simplify computations. These approximations allow us to treat complex geometries using simpler models, such as considering only the leading terms in potential contributions or equating distributed charges to point charges, which greatly simplifies the analysis of the system’s electrical properties.

Superposition Principle in Electrostatics

The superposition principle states that the total potential at any point due to multiple sources is the algebraic sum of the potentials due to each individual source. This principle is particularly useful when dealing with systems where conductors are sufficiently separated, allowing the potential on one conductor to be approximated as the sum from both charges.

Series Combination of Capacitors

When capacitors are combined in series, the total or equivalent capacitance is determined by the reciprocal of the sum of the reciprocals of the individual capacitances. In a system of two conductors, as the separation increases such that their mutual influence becomes negligible, the configuration behaves like two capacitors in series.

Capacitance

Capacitance is defined as the ratio of the charge stored on a conductor to the potential difference applied between it and another conductor. In any system, calculating the effective capacitance involves understanding how charges distribute on the conductors and how the potentials generated by these charges interact with each other.

Electric Potential of a Charged Conductor

The electric potential due to a charged conductor, such as a sphere, is determined by its geometry and the charge residing on it. For conductors with spherical symmetry, the potential outside the conductor is equivalent to that of a point charge located at the center, which simplifies the analysis when considering interactions between multiple conductors.

Transcript

00:01 So we have two spheres, and we'll give one a radius of a and one a radius b.

00:12 And they're separated by a distance d.

00:17 So we're supposed to assume that the potential of each sphere is just the sum of the potentials.

00:24 So the potential at the surface of a is going to be k, q, over a, and then we'll subtract out the potential.

00:38 From the other sphere, which is a distance d away.

00:44 And then we can do the same thing for the potential at b, except we're going in the opposite direction, so we'll just take this whole thing and take the opposite of it.

00:57 And now the capacitance of this system is going to be q divided by, well, the potential difference, so va minus vb, which let me write the denominator first.

01:17 So this is going to be kq over a minus kq over d.

01:24 And then we're subtracting that, so we'll just add the same thing.

01:31 Kq over b minus kq over d.

01:35 This is q on top.

01:38 And so we get q over, well, let's see, this is, i can factor out a kq, then i get one.

01:48 1 over a plus 1 over b minus 1 over d minus 1 over d.

01:58 So this turns into, well, the cues cancel.

02:02 1 over k, well, k is 1 over 4 pi epsilon not.

02:06 So 4 pi epsilon not goes on the numerator...