It is asking for the magnitude and direction of the electric field strength

[happyknife]

What are the magnitude and direction of the electric field strength at point Z in the following:

Point X, q1 = -2.0 x 10^-5 and it is 60 cm to the left of point Y, q2 = 8.0 x 10^-6, which is 30 cm to the left of Z.

Essentially, this is what it looks like:
X-------Y--Z

 

[Bhumble]

This is a matter of superposition. Just calculate the E field for each independent of the other charge and add.

 

[happyknife]

This is a matter of superposition. Just calculate the E field for each independent of the other charge and add.

I did, but I'm not not getting the correct answer, which should be 5.8 x 10^5. I'm not sure how they got that, which is what I'm essentially asking for: The process. I end up having to add -222 222.2222 + 8.0 x 10^5, however it's obviously not going to be the right answer.

 

[Bhumble]

$$\oint E \bullet da$$= $$\frac{Q_enc}{\epsilon_{0}}$$

$$E = \frac{Q_{enc}}{4\pi\epsilon_{0}r^2}$$$$E_x = \frac{-2.0x10^-5}{4\pi\epsilon_{0}(0.9)^2} = -2.22x10^5$$

$$E_y = \frac{8.0x10^-6}{4\pi\epsilon_{0}(0.3)^2} = 7.99x10^5$$

$$E = E_x + E_y = -2.22x10^5 + 7.99x10^5 = 5.77x10^5$$

Looks like you had the right answer already. Is there a part you don't understand or is this just a slight oversight? And direction is radially inward or outward for a given negative or positive charge. In this case just along whatever you define the axis as.

 

[happyknife]

I don't remember using that formula. I've been using the formula Electric field = Cq/r^2, C being a constant 9.0 x 10^9.

 

[musiliu]

your constant C is equal to k = 1 / 4 pi e₀, so the equations are the same..