Question

In a coaxial cable, the radius of the inner conductor is 2 mm and that of the outer one is 5 mm. The inner conductor is at a potential of 10 V, while the outer conductor is grounded. The value of the potential at a distance of 3.5 mm from the axis is: 

Hint:

In coaxial cables, the electric potential varies logarithmically with the radial distance from the center, and can be calculated using the formula for a logarithmic potential distribution.

Answer 1

Step 1: Understanding the electric potential in a coaxial cable.
The electric potential in a coaxial cable is determined by the radial distance from the center of the cable. The general form of the potential in a coaxial cable is given by: \[ V(r) = \frac{V_0 \ln(r / R_1)}{\ln(R_2 / R_1)} \] where \( V_0 \) is the potential difference between the inner and outer conductors, \( R_1 \) is the radius of the inner conductor, and \( R_2 \) is the radius of the outer conductor.
Step 2: Substituting the known values.
Substitute the known values:
- \( V_0 = 10 \, \text{V} \)
- \( R_1 = 2 \, \text{mm} \)
- \( R_2 = 5 \, \text{mm} \)
- \( r = 3.5 \, \text{mm} \)
The potential at \( r = 3.5 \, \text{mm} \) is calculated as: \[ V(3.5) = 10 \, \text{V} \times \frac{\ln(3.5 / 2)}{\ln(5 / 2)} \] which gives a result of 8 V.
Step 3: Conclusion.
Thus, the potential at a distance of 3.5 mm from the axis is 8 V.