The electric field produced by a single-point charge is given by
[tex] E(r)=k\frac{q}{r^2} [/tex]
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 [tex] q=-3.00 nC=-3.00 \cdot 10^{-9} C [/tex], and it is located at x=0, so its distance from the point x=0.200 m is
[tex] r=0.200 m-0=0.2 m [/tex]
Therefore, the electric field is
[tex] E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C [/tex]
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 [tex] q=-5.50 nC=-5.5 \cdot 10^{-9}C [/tex] and it is located at x=0.800 m, so its distance from the point is
[tex] r=0.800 m-0.200 m=0.6 m [/tex]
Therefore, the electric field is
[tex] E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N [/tex]
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
[tex] E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C [/tex]
and the sign tells us that the field is directed toward negative x-direction.
