Animated Solution for Physics - Current Electricity: Space between two concentric conducting spheres of radii $a$ and $b(b>a)$ is filled with a medium of resistivity $\rho$. The resistance between the two spheres (in ohm) will be

Imagine you are an electron standing on the surface of the inner conducting sphere of radius $a$. Your destination is the outer sphere of radius $b$. To get there, you must travel through a resistive medium with resistivity $\rho$.

At first glance, you might be tempted to reach for the trusty old formula for resistance:

But there is a catch here. This formula only works when the cross-sectional area $A$ remains perfectly constant along the entire length $l$ of the conductor. As you travel radially outwards from the inner sphere to the outer sphere, the space you are moving through is expanding. The cross-sectional area perpendicular to your path is the surface area of a sphere, which grows larger and larger the further out you go. Because the area is continuously changing, applying the standard formula directly will lead you straight into a trap.

Slicing the Sphere

When algebra fails us because things are changing, we call upon calculus. We need to break this massive, expanding journey into infinitesimally small steps where the area can be considered practically constant.

Visualize a very thin, hollow spherical shell located at a distance $x$ from the center, with an infinitesimally small thickness $dx$.

For this incredibly thin shell, the current travels a distance equal to its thickness, so the length $l$ is simply $dx$. The cross-sectional area the current passes through is the surface area of this spherical shell, which is $A=4\pi x^2$.

Now, we can safely apply our resistance formula to this tiny slice to find its elemental resistance $dR$:

Since the current must flow through each of these concentric shells one after another to reach the outer sphere, all these elemental shells are connected in series. In a series circuit, resistances simply add up. To sum up an infinite number of infinitesimally small resistances, we integrate $dR$ from the inner radius $a$ to the outer radius $b$.

Don't get intimidated by the integral. The resistivity $\rho$ and the geometric factor $4\pi$ are constants, so we can pull them outside the integral sign:

The integral of $x^{-2}$ is simply $-\frac{1}{x}$. Now, we just need to evaluate this from our lower limit $a$ to our upper limit $b$:

Substituting the limits, we get:

Notice how the double negative turns into a positive, allowing us to flip the terms for a cleaner expression. This gives us our final, elegant result:

This result beautifully captures the geometry of the radial flow. It tells us that the resistance is heavily dominated by the region near the inner sphere (where $a$ is small and the area is tightest), which is a profound physical insight!