What are the magnitude and direction of the electric field at a point midway between a -8.0 µC and a +5.0 µC charge 3.5 cm apart? Assume no other charges are nearby.
FIND MAGNITUDE:
Equation:
E=(k/r^2)(absolute value q1 + absolute value q2)(10^-6)
plug in numbers:
E=(9.0x10^9/.035^2)(8+5)(10^-6) = 95510204.08 N/C
2007-09-02 13:19:21 · 2 answers · asked by Jessie L 2 in Science & Mathematics ➔ Physics
I added -8 and 5 to get -3. I took the absolute value of that and mulitiplied to other numbers yet the result was still incorrect.
2007-09-02 14:08:58 · update #1
The formula instructs you to MULTIPLY charge magnitudes; instead, you're ADDING them. Just use the correct procedure. That's all.
2007-09-02 14:10:58 · answer #1 · answered by Jicotillo 6 · 0⤊ 0⤋
You should not take the "absolute value" of the charges first before addition, instead, you should do the addition first before take the "absolute value". This is some thing like, say, the absolute difference of points on the Yahoo-Answer. Suppose both you and me start from zero points. You gain 200, I gain -100, the absolute difference is NOT:
|200| - |-100| = 100
2007-09-02 13:31:29 · answer #2 · answered by Hahaha 7 · 0⤊ 0⤋