How to integrate the electric field of the square sheet

[garylau]

Sorry
i have one question to ask

how to integrate the electric field of the square sheet( see the pink circle below)
it looks hard for me

thank you very much

 

[blue_leaf77]

Hint: Try substitution method $2u+z^2=t$. You might still need one more substitution, but I will not comment any further before you show your own work.

 

[cnh1995]

Use the substitution
√(2u+z²)=t.

 

[blue_leaf77]

Yes you can also do that, and may be in the third line you can use the fact that the derivative of $\sec x$ is $\sec x \tan x$. But your way is kind of longer than necessary.

 

[garylau]

Yes you can also do that, and may be in the third line you can use the fact that the derivative of $\sec x$ is $\sec x \tan x$. But your way is kind of longer than necessary.

Did i make mistake in my calculation?

 

[blue_leaf77]

Did i make mistake in my calculation?

Looks good. Now you only need to do the last integral and change back to the original variable $u$ and plug in the integral limits.

 

[cnh1995]

Use the substitution
√(2u+z²)=t.

If you use this substitution,
du/√(2u+z²) can be replaced by 'dt' and u+z²=(t²+z²)/2.
So, you'll simply get it as ∫2dt/(t²+z²) which is (2/z)tan^-1(t/z).
You get your answer in just two steps.

 

[garylau]

If you use this substitution,
du/√(2u+z²) can be replaced by 'dt' and u+z²=(t²+z²)/2.
So, you'll simply get it as ∫2dt/(t²+z²) which is (2/z)tan^-1(t/z).
You get your answer in just two steps.

Oh i see thank you

 

[garylau]

Looks good. Now you only need to do the last integral and change back to the original variable $u$ and plug in the integral limits.

Looks good. Now you only need to do the last integral and change back to the original variable $u$ and plug in the integral limits.

i don't know why i do it wrong (is there a minus sign??)
can you help me to check it
thank

 

[blue_leaf77]

I missed one mistake in your work in post #5. In the last line, you should have removed the integral and the integration element. There should only be #\theta## there.

i don't know why i do it wrong (is there a minus sign??)
can you help me to check it
thank

I don't know why you are redoing your work, you are almost there in post #5.

 

[garylau]

I missed one mistake in your work in post #5. In the last line, you should have removed the integral and the integration element. There should only be #\theta## there.

I don't know why you are redoing your work, you are almost there in post #5.

i redo my work by other way

and i found the answer looks different from my answer in post 5(which i successfully do it)
something looks crazy when the answer in my last post looks totally different.
but i cannot find any mistake

 

[garylau]

I missed one mistake in your work in post #5. In the last line, you should have removed the integral and the integration element. There should only be #\theta## there.

I don't know why you are redoing your work, you are almost there in post #5.

yes

i should remove in the integral but i always forget

thank you

 

[Kshitish]

If you use this substitution,
du/√(2u+z²) can be replaced by 'dt' and u+z²=(t²+z²)/2.
So, you'll simply get it as ∫2dt/(t²+z²) which is (2/z)tan^-1(t/z).
You get your answer in just two steps.

what if i try to integrate it using multiple integration...seems quite tough then...can you help regarding that??