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

In summary, the magnitude and direction of the electric field strength at point Z can be calculated using the superposition principle. Each charge at points X and Y should be treated independently and their electric fields should be added together. The formula for electric field is E = Q_enc / 4piepsilon0r^2, where Q_enc is the enclosed charge and r is the distance from the charge to the point of interest. The result for the electric field at point Z is -5.77x10^5, with a direction determined by the sign of the charges at points X and Y. The constant C in the electric field formula is equal to 1 / 4piepsilon0.
  • #1
happyknife
4
0
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
 
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  • #2
This is a matter of superposition. Just calculate the E field for each independent of the other charge and add.
 
  • #3
Bhumble said:
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.
 
  • #4
[tex]\oint E \bullet da [/tex]= [tex]\frac{Q_enc}{\epsilon_{0}}[/tex]

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

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

[tex] E = E_x + E_y = -2.22x10^5 + 7.99x10^5 = 5.77x10^5 [/tex]

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.
 
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  • #5
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.
 
  • #6
your constant C is equal to k = 1 / 4 pi e0, so the equations are the same..
 

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

What does "magnitude" refer to in the context of electric field strength?

In the context of electric field strength, magnitude refers to the numerical value or strength of the electric field at a specific point.

What units are typically used to measure electric field strength?

Electric field strength is typically measured in units of volts per meter (V/m).

How is the direction of electric field strength represented?

The direction of electric field strength is represented by the orientation of the electric field lines. These lines point in the direction that a positively charged particle would move if placed in the field.

Can the direction of electric field strength change?

Yes, the direction of electric field strength can change depending on the location and configuration of the electric charges creating the field.

What factors can affect the magnitude of electric field strength?

The magnitude of electric field strength can be affected by the distance between charges, the strength of the charges, and the presence of other objects or materials in the vicinity.

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