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Different answer from teacher's .

  • Thread starter dncnqn
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different answer from teacher's.....

Okay, here's what the question says in my book: The electric field midway between two equal but opposite point charges is 1750 N/C, and the distance between the charges is 16.0 cm. What is the magnitude of the charge on each?
I found what the force would be equal to and put it in place of F in E=F/q. When I worked it out I got 4.98 x 10^-9 C but the answer my teacher gave us was 6.2 x 10^-10 C. So am I not getting it or did he give the wrong answer? Any help would be much appreciated, thanks!
 

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dncnqn said:
Okay, here's what the question says in my book: The electric field midway between two equal but opposite point charges is 1750 N/C, and the distance between the charges is 16.0 cm. What is the magnitude of the charge on each?
I found what the force would be equal to and put it in place of F in E=F/q. When I worked it out I got 4.98 x 10^-9 C but the answer my teacher gave us was 6.2 x 10^-10 C. So am I not getting it or did he give the wrong answer? Any help would be much appreciated, thanks!
Er.... what "force" are you calculating? The force exerted on one by the other? How could you know this? All you are given is the electric field at the midway point. There's NO charge here for you to use for the "q" in F=qE. Both of the charges given combine to produce the "E". None of them can be used as a "test charge" anymore for you to use as "q". They are both the SOURCE of the electric field.

Look up (or derive) the expression of the E-field for a point charge. You have two point charges. The electric field at the midway point for your problem is the superposition from each of them. In fact, since the point is on a line between these two, and the E-field from each points in the same direction, the E-field from the individual charges ADDS.

Zz.
 

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