# Electric field of "half" an infinite charged sheet

#### BearY

1. The problem statement, all variables and given/known data
A charged sheet with charge density $\sigma$ is described by $-\infty<x<0,-\infty<y<\infty, z = 0$. Find the electric field at $(0,0,z)$.

2. Relevant equations
Electric field of continuous density charged body from the Coulomb law:
$$E = \frac{1}{4\pi \epsilon_0}\int_V\frac{\vec{r} − \vec{r_0}}{|\vec{r} − \vec{r_0}|^3}\rho(\vec{r}')dV'$$

3. The attempt at a solution
First I tried to do a double integral but handling symmetry of x y at the same time seems very messy. So I did integral on y dimension first. It is infinite line of charge with density $\sigma dx$, and the answer is $$\frac{\sigma dx}{2\pi \epsilon_0\sqrt{x^2+z^2}}$$
Then integrate the answer from $x=-\infty$ to $x=0$.
What I did is integrate z component and x component separately, and do vector addition and take mod in the very end.
So I ended up with 2 integrals:
x component$$\frac{\sigma}{2\pi \epsilon_0}\int_{-\infty}^{0}\frac{x}{x^2+z^2}dx$$
and z component$$\frac{\sigma}{2\pi \epsilon_0}\int_{-\infty}^{0}\frac{z}{x^2+z^2}dx$$
These integral look reasonably simple, but the result I get is strange, so I guess I am wrong at some point. The first one is $\frac{ln(x^2+z^2)}{2}$, and the definite integral gives intinity. The second one is $arctan(\frac{x}{z})$ gives $\frac{\pi}{2}$.
I think this kind of question should have a finite answer and the answer given is $$\frac{\sigma}{4\epsilon_0}$$ which is finite. What went wrong?

Related Advanced Physics Homework News on Phys.org

#### BearY

Actually never mind, the question asked for z component only, I can't believe I overlooked it and spent 2 hours on this.

"Electric field of "half" an infinite charged sheet"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving