I was doing some calculations regarding the difference between the electric field generated by point charges vs a continuous line of charge (trying to come up with a cool/challenging physics problem), and I came across this interesting result which seems incorrect to me:

Imagine you have an infinite line of point charges each with charge q, separated by distance x. At one end of the line, you measure the electric field at distance x from the last charge. To visualize, the measurement point is m and the charges are c:

(m) --x-- (c) --x-- (c) --x-- (c) ...

When calculating the electric field at point m, you get: $$ \frac{kq}{x^2}+\frac{kq}{(2x)^2}+\frac{kq}{(3x)^2}+...=\frac{kq}{x^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+...\frac{1}{n^2})=\frac{kq\pi^2}{6x^2} $$

This is because $1+\frac{1}{2^2}+\frac{1}{3^2}+...\frac{1}{n^2}=\frac{\pi^2}{6}$ by some difficult proofs.

Now lets image instead of the line of point charges, you have a continuous infinite line of charge which starts a distance x away from the measurement point, with charge density q/x. To visualize, the measurement point is m:

(m) --x-- (---------Infinite line of charge with charge density q/x---------)

Now to calculate the field:

$$ \int_{x}^{\infty}\frac{kq}{xr^2}dr=-(\frac{kq}{x\infty})+(\frac{kq}{x^3})=\frac{kq}{x^3} $$

Which is clearly quite different from the above result.

I feel like I have made a simple error in either my calculations or in my conception of the problem. It seems that both situations should have the same (or at least similar) results, as both have the same linear charge density. So I guess my question is this: are my above calculations valid, and if so, is there an easy way to understand why these results are different?


1 Answer 1


You simply made a mistake in the integral:

$$\int_x^\infty \frac{kq}{xr^2} dr = \frac{kq}{x} \left.\frac{1}{r}\right|^x_\infty = \frac{kq}{x^2}.$$

You can see the mistake immediately from the fact that your result doesn't have the right units - always always check your units!

So what changes? The discrete distribution produces a field that is larger by a factor of $\pi^2/6 \approx 1.6$. This makes sense: picture dividing the continuous line into segments of length $x$ and charge $q$. Going from this to the discrete distribution means concentrating all the charge of each segment at its left endpoint, thus moving it closer to the measurement point and strengthening the field.

  • $\begingroup$ Thanks, I felt I must have messed up the integral somewhere but I can't believe I missed such an error! $\endgroup$
    – Sam
    Commented Jan 6, 2022 at 16:57

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