Answer:
(a). The potential is 78.0 kV.
(b). The potential is zero.
Step-by-step explanation:
Given that,
Radius = 6.00
Charge density = 8.50 μC/m
(a). The surface of the cylinder and 4.00 cm away from the surface,
We need to calculate the voltage
Using formula of potential
![V=(\lambda)/(2\pi\epsilon_(0))ln((r)/(R))](https://img.qammunity.org/2021/formulas/physics/college/lpwf0cllz661s1jhagdw9o8n4q9agqs97w.png)
Put the value into the formula
![V=(8.5*10^(-6))/(2\pi*8.85*10^(-12))ln((6+4)/(6))](https://img.qammunity.org/2021/formulas/physics/college/83qfdwknq2tjt3c8m37alam2er6lr3m1z7.png)
![V=78.0\ kV](https://img.qammunity.org/2021/formulas/physics/college/aluvowygtg0nh2fvzpy08lg4km6rxt32ez.png)
(b). The surface and a point 1.00 cm from the central axis of the cylinder
Here, r = R
We need to calculate the voltage
Using formula of potential
![V=(\lambda)/(2\pi\epsilon_(0))ln((r)/(R))](https://img.qammunity.org/2021/formulas/physics/college/lpwf0cllz661s1jhagdw9o8n4q9agqs97w.png)
Put the value into the formula
![V=(8.5*10^(-6))/(2\pi*8.85*10^(-12))ln(1)](https://img.qammunity.org/2021/formulas/physics/college/prxda16d1uf3016crricn67v1jlkv7nezx.png)
![V=0\ kV](https://img.qammunity.org/2021/formulas/physics/college/529njuw2e87y3r41ddkjvbs0pfzcp6iy1d.png)
Hence, (a). The potential is 78.0 kV.
(b). The potential is zero.