In the figure below, d=7.2 cm. What are the magnitude and direction of the elect
ID: 3898873 • Letter: I
Question
In the figure below, d=7.2 cm. What are the magnitude and direction of the electric field at the position indicated by the dot (*) in the figure? (I can't get the figure to post - Hope my diagram works)
10nC +............d............ - 5nC
. .
. .
2d .
. .
. .
5nC + .........................*
I received almost all my points for the problem and the magnitude is 1.0 x 10^4 N/C and the direction should be 5.2 degrees below the horizontal. I am not sure if my missing points are due to not originally getting the direction correct or maybe some due to my work and maybe getting some Qx and Qy's signs incorrect but still coming up with the correct answer by luck. I labeled 5nC Q1, 10nC Q2, -5nC Q3. For Q1x I got 8670.9 N/C, Q3y -2167.7 N/C, Q2x 1551.3 N/C and Q2y 3101.9 N/C.
If you could work out the problem I would appreciate it to compare to my work. Thanks!
Explanation / Answer
Enet = E1 + E2 + E3
E = kq/r^2
E1 = 9*10^9*5*10^-9 / 0.072^2 = 8680.55 i N/C
E2 = 9*10^9*10*10^-9 / (5*0.072^2) * cos(theta) i + sin(theta) j
theta = arctan(2) = 63.43 deg
E2 = 3472.22 * (cos(63.43 deg) i - sin(63.43 deg) j)
E2 = 1553.09 i - 3105.51 j N/C
E3 = 9*10^9*5*10^-9 / (4*0.072^2) = 2170.1389 j N/C
Enet = i(8680.55+1553.09) + j(2170.1389-3105.51) = 10233.64 i - 935.37 j N/C
Magnitude = sqrt(10233.64^2 + 935.37^2) = 10276.298 N/C
Direction = arctan(-935.37/10233.64) = 360 - 5.22 = 354.78 degrees