The electric field produced by a single-point charge is given by
![E(r)=k\frac{q}{r^2}](https://tex.z-dn.net/?f=%20E%28r%29%3Dk%5Cfrac%7Bq%7D%7Br%5E2%7D%20%20)
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.
1) The first charge is
, and it is located at x=0, so its distance from the point x=0.200 m is
![r=0.200 m-0=0.2 m](https://tex.z-dn.net/?f=%20r%3D0.200%20m-0%3D0.2%20m%20)
Therefore, the electric field is
![E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C](https://tex.z-dn.net/?f=%20E_1%3D%288.99%20%5Ccdot%2010%5E9%20Nm%5E2C%5E%7B-2%7D%29%5Cfrac%7B%283.0%20%5Ccdot%2010%5E%7B-9%7D%20C%29%7D%7B%280.2%20m%29%5E2%7D%3D675%20N%2FC%20%20)
And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.
2) The second charge is
and it is located at x=0.800 m, so its distance from the point is
![r=0.800 m-0.200 m=0.6 m](https://tex.z-dn.net/?f=%20r%3D0.800%20m-0.200%20m%3D0.6%20m%20)
Therefore, the electric field is
![E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N](https://tex.z-dn.net/?f=%20E_2%20%3D%20%288.99%20%5Ccdot%2010%5E9%20Nm%5E2C%5E%7B-2%7D%29%5Cfrac%7B%285.5%20%5Ccdot%2010%5E%7B-9%7D%20C%29%7D%7B%280.6%20m%29%5E2%7D%3D137.5%20N%20%20)
And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.
3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have
![E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C](https://tex.z-dn.net/?f=%20E%3DE_2%20-E_1%20%3D137.5%20N%2FC%2FC-675%20N%2FC%3D-537.5%20N%2FC%20)
and the sign tells us that the field is directed toward negative x-direction.