- #1

lampCable

- 22

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## Homework Statement

Solve ##\Delta\phi = -q\delta(x)## on ##\mathbb{R}##.

Correct answer: ##\phi = -\frac{q}{2}|x| + Ax + B##

## Homework Equations

## The Attempt at a Solution

In one dimension the equation becomes ##\frac{d^2 \phi}{d x^2} = -q\delta(x)##. We integrate from ##-\infty## to ##x## to obtain

$$\int_{-\infty}^x \frac{d^2 \phi}{d x^2} dx = \frac{d \phi}{dx}(x) - \frac{d \phi}{dx}(-\infty) = \int_{-\infty}^x -q\delta(x) dx.$$

If ##x<0##, the last integral becomes ##0##, from which it follows that ##\frac{d \phi}{dx}(x) = \frac{d \phi}{dx}(-\infty)##. Since this holds for all ##x<0## we find that ##\frac{d \phi}{dx}(-\infty) = A = \text{constant}##.

If ##x>0##, the last integral becomes ##-q##, so we find that ##\frac{d \phi}{dx}(x) = -q + \frac{d \phi}{dx}(-\infty) = -q + A##.

The same procedure with an integral from ##x## to ##\infty## we find that for ##x<0## the potential is ##\frac{d \phi}{dx}(x) = q + B##, where ##C## is some other constant.

(1) Where do I go from here? My idea is that I need to connect the two parts and identify that ##\frac{d \phi}{dx}(x) = -\frac{q}{2}\text{sign}(x) + C##, where ##C## some constant.

(2) Are there any other approaches to this problem?