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
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.
1 answer
1 answer