Electric Field at Midway Point of Two Opposite Charges

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SUMMARY

The discussion centers on calculating the electric field at the midpoint between two equal but opposite charges of 7e-07 C, positioned 0.7 m apart on the x-axis. Initially, the assumption was that the electric fields would cancel each other out, resulting in a net electric field of zero. However, it was concluded that the electric field at the midpoint is actually doubled due to the opposing nature of the charges, with the electric field pointing away from the positive charge and toward the negative charge.

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GenMipps
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This problem looked so easy at first:

Two opposite charges of equal magnitude 7e-07 C are held on an x-axis 0.7 m apart, with the negative charge on the right. What are the magnitude and direction of E at the point midway between charges?

I had the idea that I would solve this by summing up the electric fields.
Et = E1 + E2
and
E = kq/r^2

I started to do the numbers, then stopped when it become very obvious that since the two charges were equal but opposite in charge, the electric field midway would have to be zero, as we would have two equal electric fields cancelling each other out. But the online homework marker says it isn't zero. What part of my assumption is wrong?
 
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No, the electric field points away from + charges and toward - charges. The E field accelerates + charges in the direction of the E field vector. Does that help? What equation specifies the electric field distribution due to a point charge?
 
Are you sure the electric fields cancel each other out? You may want to review their directions.
 
Wow, I am an idiot, I solved it myself. For anyone who cares, the fields would cancel each other out if they were equal. Since they are opposite, the field in the middle is doubled.
 
No, you're not an idiot. We've all made mistakes like those. At least you realized that the result didn't seem right. Welcome to PF, BTW.:biggrin:
 

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