What is the electric field at x=2d?

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SUMMARY

The electric field at x=2d, due to point charges q at x=0 and 2q at x=3d, is calculated using the formula E=kQ/r^2. The contributions to the electric field are E1=kq/(2d)^2 resulting in kq/4d^2 from charge q, and E2=2kq/d^2 from charge 2q, leading to a net electric field of E1-E2=7kq/4d^2 directed to the left. However, the correct answer should be expressed as -7kq/4d^2, indicating the direction is negative, which was misinterpreted in the formatting required by the online homework system.

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Homework Statement


Positive point charges q and 2q are located at x=0 and x=3d respectively. What is the electric field at x=2d?

Homework Equations



E=kQ/r^2

The Attempt at a Solution


E1=kq/(2d)^2=kq/4d^2
E2=2kq/d^2
E1-E2=7kq/4d^2, to the left

To the left is correct but the answer is wrong :( can someone please tell me why? thanks.
 
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It should be the negative of that. Small mistake but you did everything else correctly.
-7kq/4d^2
 
0ddbio said:
It should be the negative of that. Small mistake but you did everything else correctly.
-7kq/4d^2

When I said left, that was intended to mean the sign was negative. In my online homework, for vectors we can not put in actual signs. For example we have to put it in this format:

magnitude of vector, direction.

that is why i said 7kq/4d^2, to the left--assuming to the right is positive and to the left is negative.

but it's still saying that is wrong :(
 
You're work is right; either the format is wrong, or the program has the wrong solution.

Are you sure you're not supposed to plug in for k or something?
 
What is the answer that you're trying to match?
 

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