How Do You Determine the Direction of an Electric Field?

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SUMMARY

The discussion focuses on determining the direction of an electric field acting on a negative charge of -5.0C with a force of 2.5N directed to the left. The electric field (ε) is calculated as 0.5N/C using the formula ε = FE/q. The confusion arises regarding the direction of the electric field, with clarification that a negative charge experiences a force in the opposite direction of the electric field. Thus, the charge will move to the right, opposite to the direction of the applied force.

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  • Proficiency in using the equation F = qE
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Balsam
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Homework Statement


An electric field with a magnitude of 2.5N directed to the left, acts on a negative charge of -5.0C. Determine the electric field in which the charge is located.

Give. FE=2.5N
q=-5.0C

Homework Equations


FE[/B]=qε

The Attempt at a Solution


ε[/B]=2.5N/5.0C=0.5N/C. I don't know how to find the direction. The answer is 0.5N/C,
but I don't know how to get that direction. I guess we have to assume the positive test charge is to the right of q and figure out the direction it would move as a result of FE applied on it by q. That doesn't really make sense though.

Edit: I don't know why the formatting is weird. Also, please help- my exam is tomorrow!​
 
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Use vector equation ##\mathbf F = q \mathbf E##. With ##\mathbf F = -2.5 \hat x## and ##q = -5\hspace{2mm} C##, solve for ##\mathbf E##.
 
blue_leaf77 said:
Use vector equation ##\mathbf F = q \mathbf E##. With ##\mathbf F = -2.5 \hat x## and ##q = -5\hspace{2mm} C##, solve for ##\mathbf E##.
I did, but I don't know how to find the direction
 
You did, then please write out your answer for ##\mathbf E##, note that the bold symbol denotes vector notation so your answer must consist of a number and the unit vector.
 
blue_leaf77 said:
You did, then please write out your answer for ##\mathbf E##, note that the bold symbol denotes vector notation so your answer must consist of a number and the unit vector.
Ok, and how do you find the direction?
 
If you really have calculated ##\mathbf E##, not just ##|\mathbf E|##, then show it in your next comment.
 
blue_leaf77 said:
If you really have calculated ##\mathbf E##, not just ##|\mathbf E|##, then show it in your next comment.

My teacher said you're not supposed to plug in - signs into the equations for fields because it implies direction, so we only use magnitudes
 
I don't know in which manner your teacher wants you to find the direction, but the formal way is to use the vector equation. Alternatively, you can use the definition of negative charge. A negative charge is defined to be the condition such that the field the particle is exposed to exert a force in an opposite direction to that field.
 
blue_leaf77 said:
I don't know in which manner your teacher wants you to find the direction, but the formal way is to use the vector equation. Alternatively, you can use the definition of negative charge. A negative charge is defined to be the condition such that the field the particle is exposed to exert a force in an opposite direction to that field.

I'm confused about the wording of the question. Usually we use a positive test charge along with the given charge. Is the test charge applying the force on q? Then it would be to the right of q since the force is directed to the left. And then q would move to the right as it is attracted by the positive test charge?
 
  • #10
Balsam said:
Then it would be to the right of q since the force is directed to the left.
That's indeed the definition of negative charge when applied to the current problem at hand.
Balsam said:
And then q would move to the right as it is attracted by the positive test charge?
No, the charge will move in the same direction as the force, not the field.
 
  • #11
blue_leaf77 said:
That's indeed the definition of negative charge when applied to the current problem at hand.

No, the charge will move in the same direction as the force, not the field.
But the answer is
 
  • #12
Balsam said:
Then it would be to the right of q since the force is directed to the left
You already arrived at the correct answer there.
 

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