Answer:
The magnitude of the electric field and direction of electric field are
and 75.36°.
Explanation:
Given that,
First charge ![q_{1}= 2.9\mu C](https://tex.z-dn.net/?f=q_%7B1%7D%3D%202.9%5Cmu%20C)
Second charge![q_{2}= 2.9\mu C](https://tex.z-dn.net/?f=q_%7B2%7D%3D%202.9%5Cmu%20C)
Distance between two corners r= 50 cm
We need to calculate the electric field due to other charges at one corner
For E₁
Using formula of electric field
![E_{1}=\dfrac{kq}{r'^2}](https://tex.z-dn.net/?f=E_%7B1%7D%3D%5Cdfrac%7Bkq%7D%7Br%27%5E2%7D)
Put the value into the formula
![E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}](https://tex.z-dn.net/?f=E_%7B1%7D%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes2.9%5Ctimes10%5E%7B-6%7D%7D%7B%2850%5Csqrt%7B2%7D%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![E_{1}=52200=52.2\times10^{3}\ N/C](https://tex.z-dn.net/?f=E_%7B1%7D%3D52200%3D52.2%5Ctimes10%5E%7B3%7D%5C%20N%2FC)
For E₂,
Using formula of electric field
![E_{1}=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E_%7B1%7D%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
Put the value into the formula
![E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}](https://tex.z-dn.net/?f=E_%7B2%7D%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes2.9%5Ctimes10%5E%7B-6%7D%7D%7B%2850%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![E_{2}=104400=104.4\times10^{3}\ N/C](https://tex.z-dn.net/?f=E_%7B2%7D%3D104400%3D104.4%5Ctimes10%5E%7B3%7D%5C%20N%2FC)
We need to calculate the horizontal electric field
![E_{x}=E_{1}\cos\theta](https://tex.z-dn.net/?f=E_%7Bx%7D%3DE_%7B1%7D%5Ccos%5Ctheta)
![E_{x}=52.2\times10^{3}\times\cos45](https://tex.z-dn.net/?f=E_%7Bx%7D%3D52.2%5Ctimes10%5E%7B3%7D%5Ctimes%5Ccos45)
![E_{x}=36910.97=36.9\times10^{3}\ N/C](https://tex.z-dn.net/?f=E_%7Bx%7D%3D36910.97%3D36.9%5Ctimes10%5E%7B3%7D%5C%20N%2FC)
We need to calculate the vertical electric field
![E_{y}=E_{2}+E_{1}\sin\theta](https://tex.z-dn.net/?f=E_%7By%7D%3DE_%7B2%7D%2BE_%7B1%7D%5Csin%5Ctheta)
![E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45](https://tex.z-dn.net/?f=E_%7By%7D%3D104.4%5Ctimes10%5E%7B3%7D%2B52.2%5Ctimes10%5E%7B3%7D%5Csin45)
![E_{y}=141310.97=141.3\times10^{3}\ N/C](https://tex.z-dn.net/?f=E_%7By%7D%3D141310.97%3D141.3%5Ctimes10%5E%7B3%7D%5C%20N%2FC)
We need to calculate the net electric field
![E_{net}=\sqrt{E_{x}^2+E_{y}^2}](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D%5Csqrt%7BE_%7Bx%7D%5E2%2BE_%7By%7D%5E2%7D)
Put the value into the formula
![E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D%5Csqrt%7B%2836.9%5Ctimes10%5E%7B3%7D%29%5E2%2B%28141.3%5Ctimes10%5E%7B3%7D%29%5E2%7D)
![E_{net}=146038.69\ N/C](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D146038.69%5C%20N%2FC)
![E_{net}=146.03\times10^{3}\ N/C](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D146.03%5Ctimes10%5E%7B3%7D%5C%20N%2FC)
We need to calculate the direction of electric field
Using formula of direction
![\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac%7B141.3%5Ctimes10%5E%7B3%7D%7D%7B36.9%5Ctimes10%5E%7B3%7D%7D)
![\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})](https://tex.z-dn.net/?f=%5Ctheta%3D%5Ctan%5E%7B-1%7D%28%5Cdfrac%7B141.3%5Ctimes10%5E%7B3%7D%7D%7B36.9%5Ctimes10%5E%7B3%7D%7D%29)
![\theta=75.36^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D75.36%5E%7B%5Ccirc%7D)
Hence, The magnitude of the electric field and direction of electric field are
and 75.36°.