Answer:
Answer is explained in the explanation section below.
Explanation:
Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.
Solution:
The relation between electric field and the electric potential is:
E = [tex]\frac{dV}{dr}[/tex]
So, making dV the subject, we have:
dV = E x dr
Integrating the above equation, we get.
V = [tex]\int\limits^_ {} \,[/tex]E x dr Equation 1
Now, in 2-D
E is inversely proportional to the radius r.
E ∝ 1/r
So, we can write: replacing E ∝ 1/r in the equation 1
V ∝ [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr
Which implies that,
V ∝ log (r)
Hence, distance dependence expected for the electric potential = ln (r)