# Confusion in calculating electric field due to infinite plane

(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.

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:

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

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

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

Therefore:

$$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

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 ($$E_x=-2 \pi\ k\ \sigma$$)

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