Both E are in the direction to the right. So they add.
Etotal= kqleft/(.065-.031)^2 + kqright/(.094-065)^2
You can me fancy to let the sign do the direction, as in here...
E=kq1 *(.065-.031)/ABS(.065-.031)^3 + kq2(.065-.093)/ABS(.065-.093)^3
That again, will yield a E that is +,to the right.
Two charges are placed on the x axis. One of the charges (q1 = +9.4 µC) is at x1 = +3.1 cm and the other (q2 = -22 µC) is at x2 = +9.3 cm.
Find the net electric field (magnitude and direction) at x = +6.5 cm. (Use the sign of your answer to indicate the direction along the x-axis.)
I changed the radius by 6.50-3.1 and then 6.5-9.3 and but then in the equation and then added but im not getting the right answer.
1 answer
1 answer