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Question 2.22: Find the potential a distance s from an infinitely long stra...

Find the potential a distance s from an infinitely long straight wire that carries a uniform line charge λ. Compute the gradient of your potential, and check that it yields the correct field.

Question Data is a breakdown of the data given in the question above.
  • An infinitely long straight wire
  • The wire carries a uniform line charge λ
  • The distance from the wire is s
The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
Step 1:
we set the reference point for the electric potential at a distance "a" from the charge. This is because the charge itself extends to infinity, so we cannot set the reference point at infinity.
Step 2:
we calculate the electric potential "V" at a distance "s" from the charge by integrating the electric field expression over the range from "a" to "s". This gives us the negative of the integral of 2λ/(4πε₀s) with respect to "s".
Step 3:
we simplify the integral to obtain the expression -2λln(s/a)/(4πε₀), where "ln" denotes the natural logarithm. This is the formula for the electric potential at a distance "s" from the charge.
Step 4:
we calculate the gradient of the electric potential using the formula for the gradient in spherical coordinates. Taking the derivative of the electric potential with respect to "s", we obtain -2λ/(4πε₀s), which is equal to the negative of the electric field "E".
Therefore, the negative gradient of the electric potential gives us the electric field vector, as expected from the theory.

Final Answer

E=\frac{1}{4\pi \varepsilon _0}\frac{2\lambda }{s}\hat{s} (Prob. 2.13). In this case we cannot set the reference point at ∞, since the charge itself extends to ∞. Let’s set it at s = a. Then

V (s) = −\int_{a}^{s}{}\left(\frac{1}{4\pi \varepsilon _0}\frac{2\lambda }{\bar{s} } \right) d\bar{s}=-\frac{1}{4\pi \varepsilon _0}2\lambda \ln \left(\frac{s}{a} \right) .

(In this form it is clear why a = ∞ would be no good—likewise the other “natural” point, a = 0.)

∇V=-\frac{1}{4\pi \varepsilon _0}2\lambda \frac{\partial}{\partial s} \left(\ln \left(\frac{s}{a} \right) \right)\hat{s}=-\frac{1}{4\pi \varepsilon _0}2\lambda \frac{1}{s}\hat{s} =-E .

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