Posted by **alex** on Tuesday, October 2, 2012 at 8:41pm.

Please help!

Two charges are placed on the x axis. One of the charges (q1 = +8.71C) is at x1 = +3.00 cm and the other (q2 = -28.6C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.

- physics -
**Jennifer**, Wednesday, October 3, 2012 at 1:18pm
E = k * q/r^2

The distances all have to be converted to meters because you want everything in SI units

a)

at x = 0, the field from the first charge is k * 8.71 / 0.03m^2

the field from the 2nd charge is k * -28.6 / 0.09m^2

So the total electric field is the sum of these numbers

k * 8.71 / 0.03^2 + -28.6 / 0.09^2

b) at x = 6.00 cm, the first charge is a distance of (6-3) = 3 cm away = 0.03 m, and the 2nd charge is at a distance

(9-6) = 3 cm = 0.03 m

So the total electric field is

k * 8.71 / 0.03^2 + k * -28.6 / 0.03^2

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