Sphere of inner radius a and outer radius b is made of `p` uniform resistivity find resistance between inner and outer surface

Options

A. `(rho)/(4pi)((1)/(a)-(1)/(b))`
B. `(rho)/(2pi)((1)/(a)-(1)/(b))`
C. `(rho)/(2pi)((1)/(a)+(1)/(a)+(1)/(b))`
D. `(rho)/(4pi)((1)/(a)+(1)/(b))`

Answer

To find the resistance between the inner and outer surfaces of a spherical shell with inner radius \( a \) and outer radius \( b \), made of a material with uniform resistivity \( \rho \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Resistance**:
The resistance \( R \) of a conductor is given by the formula:
\[ R = \frac{\rho l}{A} \]
where \( \rho \) is the resistivity, \( l \) is the length (or thickness) through which current flows, and \( A \) is the cross-sectional area.

2. **Consider a Thin Spherical Shell**:
To find the total resistance between the inner and outer surfaces, we consider a thin spherical shell of radius \( r \) and thickness \( dr \). The resistance of this thin shell can be expressed as:
\[ dR = \frac{\rho \, dr}{A} \]
where \( A \) is the surface area of the thin shell.

3. **Calculate the Surface Area**:
The surface area \( A \) of a sphere of radius \( r \) is given by:
\[ A = 4\pi r^2 \]
Therefore, we can substitute this into our expression for \( dR \):
\[ dR = \frac{\rho \, dr}{4\pi r^2} \]

4. **Integrate to Find Total Resistance**:
To find the total resistance \( R \) from the inner radius \( a \) to the outer radius \( b \), we need to integrate \( dR \) from \( r = a \) to \( r = b \):
\[ R = \int_{a}^{b} dR = \int_{a}^{b} \frac{\rho \, dr}{4\pi r^2} \]

5. **Perform the Integration**:
The integral can be solved as follows:
\[ R = \frac{\rho}{4\pi} \int_{a}^{b} \frac{dr}{r^2} \]
The integral of \( \frac{1}{r^2} \) is:
\[ -\frac{1}{r} \]
Thus, we have:
\[ R = \frac{\rho}{4\pi} \left[-\frac{1}{r}\right]_{a}^{b} = \frac{\rho}{4\pi} \left(-\frac{1}{b} + \frac{1}{a}\right) \]

6. **Simplify the Result**:
This simplifies to:
\[ R = \frac{\rho}{4\pi} \left(\frac{1}{a} - \frac{1}{b}\right) \]

### Final Result:
The resistance between the inner and outer surfaces of the spherical shell is:
\[ R = \frac{\rho}{4\pi} \left(\frac{1}{a} - \frac{1}{b}\right) \]