Stuck on electric field question

In summary, this person is trying to figure out how the electric field at a point due to three charges (two of which are positive and one which is negative) is calculated. They make a mistake in their calculation and need help from the other participants in the conversation.
  • #1
evangelic04
4
0
I've been stuck on this relatively simple question for over an hour, can't get the right answer. I think my fundamental understanding of the units involved is flawed, I would really appreciate any kind of clarification as to what I'm doing... okay here goes.
I'm given that three charges are at the vertices of an equilateral triangle, with side lengths of 0.8 meters. The charges are given in units of (mu)C, which I understand to be microCoulombs, is that correct? Two of the charges, mainly the left and uppermost ones, are positive; they are +2.4 microCoulombs and +6.8 microCoulombs, respectively. The third charge, on the right, is negative, a -4.2 microCoulombs. I am asked to calculate the electric field at the point of the leftmost (+2.4) charge, due to the other two charges. The answer is expected in vector format, and in kN/C (killoNewtons per Coulomb?).
This shouldn't be hard, using the equation for electric field,
Ke*(Q/r^2). What I do every time is:

convert the microCoulombs to Coulombs by dividing by 1.6 * 10^6.
So I have 6.8 * 10 ^ -6 Couloumbs, -4.9 * 10 ^ - 6 Couloumbs, and the third point doesn't matter, since that's where I'm calculating electric field at. Basically then [ Ke * (6.8 * 10 ^ -6 C) / (0.8 m) ^ 2 ], that's the magnitude, and I divide it into x and y components by multiplying by cos(60) and sin(60) respectively. x should be negative (it extends outward from positive charge to the left on the x-axis) and y is also negative (it extends outward from positive charge down on the y-axis).
For the other one, it's [ Ke * (-4.9 * 10 ^ -6 C) / (0.8 m)^2]. It lays directly horizontal to the point in question, so no y-component, it's all x-component and it's positive since the negative charge produces a field directed toward itself.
Add components, which should produce and answer in C/N, then divide by 1000 to get kN/C.
Seems simple, what am I doing wrong? Sorry if the description is too in-depth or not enough so. Please help me someone!
 
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  • #2
evangelic04 said:
What I do every time is:
convert the microCoulombs to Coulombs by dividing by 1.6 * 10^6.
"micro" just means 1 millionth. So 1 microC = 10^-6 C.
 
  • #3
I hope you're using "Ke" as a symbol for Coulomb's constant
"Ke = 9E9 [Nm^2/C^2]" , NOT as "k*e" with e=1.6E-19 [C]
 
  • #4
evangelic04, See attachments
 

Attachments

  • three charges.bmp
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  • #5
thank you all

Forgot to come back and thank you all! How rude of me!
Of course the division by 10^-6 was a bonehead error on my part, once I got past that I almost got the problem...looking at andrevdh's generous response I could see that I was getting the y-component correct everytime, it was my x-component that was wrong...I was using sin and cos on the overall magnitude, but I don't think I was fully grasping the math behind what I was trying to figure out. Anyways thanks everyone!
 

1. What is an electric field?

An electric field is a region around a charged particle or object where an electric force is exerted on other charged particles or objects.

2. How do you calculate the strength of an electric field?

The strength of an electric field is calculated by dividing the electric force on an object by the charge of the object. It is measured in units of volts per meter (V/m).

3. What causes an electric field to form?

An electric field is formed when there is a difference in electric potential between two points. This difference can be created by charged particles or objects.

4. How does an electric field affect charged particles?

An electric field exerts a force on charged particles, causing them to accelerate or move in a certain direction depending on the direction of the electric field. The magnitude of the force is determined by the strength of the electric field.

5. What are some real-world applications of electric fields?

Electric fields have many practical applications, such as in electronic devices, power transmission, and medical equipment. They are also used in technologies like particle accelerators and electrostatic precipitators.

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