(I) Electric field at a point on positive $x$-axis:

Let us consider Cartesian coordinate system with infinitely large circular plane at $y$-$z$ plane. Let $P$ be any point where we want to measure $\vec{E}$. Let us take the origin of our coordinate system at point $O$ which is the projection of point $P$ on $y$-$z$ plane.

Let us first consider a red ring of infinitesimal thickness $dl$ with center $O$ and radius $l$ in the $y$-$z$ plane.

enter image description here

Let an infinitesimal element of this red ring be $Q_1$. Let another infinitesimal element at the other end of the diameter be $Q_2$.

Let us first calculate $E_x$

$E_x$ at point $P$ due to element area at $Q_1$:

$$d^2E_x=k\ dq'\ \dfrac{\xi}{r^3}=k\ \sigma\ dS'\ \dfrac{\xi}{r^3}=k\ \sigma\ (l\ d\theta\ dl)\ \dfrac{\xi}{r^3}$$

$E_x$ at point $P$ due to element area of red ring:

$$dE_x= \int_0^{2 \pi}d^2E_x=\int_0^{2 \pi} k\ \sigma\ (l\ d\theta\ dl)\ \dfrac{\xi}{r^3}=k\ \sigma\ l\ dl\ \dfrac{\xi}{r^3} \int_0^{2 \pi} d\theta=2 \pi\ k\ \sigma\ l\ dl\ \dfrac{\xi}{r^3}$$

$E_x$ at point $P$ due to infinitely large circular plane:

$$E_x=\int_0^\infty dE_x=\int_0^\infty 2 \pi\ k\ \sigma\ l\ dl\ \dfrac{\xi}{r^3} =2 \pi\ k\ \sigma\ \xi\ \int_0^\infty \dfrac{l}{r^3}\ dl$$

$$ \bbox[5px,border:2px solid black] { \dfrac{\partial r}{\partial l}=\dfrac{dr}{dr^2} \dfrac{\partial r^2}{\partial l}=\dfrac{1}{2r} \dfrac{\partial (\xi^2 + l^2)}{\partial l}=\dfrac{1}{2r} \dfrac{\partial l^2}{\partial l}=\dfrac{1}{2r} 2l=\dfrac{l}{r} \Rightarrow dl=\dfrac{r\ dr}{l} } $$

$$ \bbox[5px,border:2px solid black] { \text{When:}\ l=0, r= \xi;\ l=\infty, r= \infty } $$


$$E_x=2 \pi\ k\ \sigma\ \xi\ \int_\xi^\infty \dfrac{l}{r^3}\ \dfrac{r\ dr}{l}= 2 \pi\ k\ \sigma\ \xi\ \int_\xi^\infty r^{-2} dr= 2 \pi\ k\ \sigma\ \xi\ \left[ -\dfrac{1}{r} \right]_{\xi}^\infty=-2 \pi\ k\ \sigma\ \xi\ \left[ \dfrac{1}{\infty}-\dfrac{1}{\xi} \right]=2 \pi\ k\ \sigma\ $$

This result is indeed correct.

(II) Electric field at a point on negative $x$-axis

enter image description here

If I use same coordinte system and find $E_x$ at a point on negative $x$-axis, I should get $(-2 \pi\ k\ \sigma)$. But I am getting $(2 \pi\ k\ \sigma)$ by following the exact same calculation as shown above. The only difference between the two calculations is that in the former $\xi$ is positive while in the latter $\xi$ is negative. But it shouldn't matter as it gets cancelled out.

Please explain why am I not getting ($E_x=-2 \pi\ k\ \sigma$)


2 Answers 2


Just looking at it summarily, it's not quite $ξ$ cancelling with itself, but rather $ξ$ cancelling with $\sqrt{ξ²}$. This is clearly seen if you keep the integration variable as $l$ rather than changing it to $r$.

$$E = k\int_0^∞(ξ²+l²)⁻¹.ξ.√(ξ²+l²)⁻¹.2πσl\ \mathrm{d}l$$


Ah, sign errors. You've got to be careful with how you treat signs.

Your issue comes from blending together coordinates and distances in $ \xi $. Coordinates may be negative, but distances may not be.

In particular, re-visit the line "When: $ l=0$, $r=\xi$." Earlier, $\xi$ referred to your coordinate along the x-axis, allowing it to be negative. In this line, however, $r=\xi$ refers to the distance from the plane, and would be non-physical for negative $\xi$. If you want $\xi$ to represent a coordinate, then this line should instead read $r=|\xi|$. The correct result follows.


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