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 potential is a function of the logarithmic relationship between the radial distance and the conductors' radii. Always ensure units are consistent when using the formula.

Answer 1

Correct Answer: 3.79

Step 1: Use the potential formula for coaxial cables.
The potential \( V(r) \) at a radial distance \( r \) from the axis in a coaxial cable is given by: \[ V(r) = \frac{1}{2\pi \epsilon_0} \ln \left( \frac{R_2}{R_1} \right) \cdot \ln \left( \frac{r}{R_1} \right) \] where \( R_1 \) and \( R_2 \) are the radii of the inner and outer conductors respectively, and \( r \) is the point where the potential is to be calculated. Step 2: Apply the given values.
Here, \( R_1 = 2 \, \text{mm} \), \( R_2 = 5 \, \text{mm} \), and \( r = 3.5 \, \text{mm} \), with the inner conductor at 10 V and the outer conductor grounded. Step 3: Calculation of potential.
By applying the formula and solving for the potential at \( r = 3.5 \, \text{mm} \), the calculated value of the potential is between 3.79 and 3.99 V.