Answer:
a) 0N/C
b) 2660.7N/C
c) 543.7N/C
Explanation:
To find the electric field in all these cases you take into account the x and y component of the total Electric field for each point.
a)
for P(0,0), only there is the x component of the field because the point is the parallel line that connects both charges:
( 1 )
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
r1: distance to the first charge = 0.150m
r:2 distance to the second charge = 0.150m
Due to in this case the distance r1=r2: you obtain, by replacing in (1):
![E_x=0](https://tex.z-dn.net/?f=E_x%3D0)
b)
for P(0.300 , 0 ) you also have only the x component of E, and the electric field generated by each charge are directed toward right:
![\vec{E}=E_x\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3DE_x%5Chat%7Bi%7D)
![E_x=k\frac{q}{r_1^2}+k\frac{q}{r_2^2}](https://tex.z-dn.net/?f=E_x%3Dk%5Cfrac%7Bq%7D%7Br_1%5E2%7D%2Bk%5Cfrac%7Bq%7D%7Br_2%5E2%7D)
r1 = 0.300m+0.150m = 0.450m
r2 = 0.300m-0.150m = 0.150m
By replacing r1 and r2 in ( 1 ) you obtain:
![E_{x}=kq(\frac{1}{r_1^2}+\frac{1}{r_2^2})=(8.98*10^9Nm^2/C^2)(6.00*10^{-9})(\frac{1}{(0.450m)^2}+\frac{1}{(0.150m)^2})\\\\E_x=2660.7N/C\\\\\vec{E}=2660.7N/C\ \hat{i}](https://tex.z-dn.net/?f=E_%7Bx%7D%3Dkq%28%5Cfrac%7B1%7D%7Br_1%5E2%7D%2B%5Cfrac%7B1%7D%7Br_2%5E2%7D%29%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%286.00%2A10%5E%7B-9%7D%29%28%5Cfrac%7B1%7D%7B%280.450m%29%5E2%7D%2B%5Cfrac%7B1%7D%7B%280.150m%29%5E2%7D%29%5C%5C%5C%5CE_x%3D2660.7N%2FC%5C%5C%5C%5C%5Cvec%7BE%7D%3D2660.7N%2FC%5C%20%5Chat%7Bi%7D)
c)
for P(0.150 , -0.40) you have both x and y components for E:
![\vec{E}=E_x\hat{i}+E_y\hat{j}\\\\E_x=k\frac{q}{r_1^2}cos\theta+0N/C\\\\E_y=-k\frac{q}{r_1^2}sin\theta-k\frac{q}{r_2^2}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3DE_x%5Chat%7Bi%7D%2BE_y%5Chat%7Bj%7D%5C%5C%5C%5CE_x%3Dk%5Cfrac%7Bq%7D%7Br_1%5E2%7Dcos%5Ctheta%2B0N%2FC%5C%5C%5C%5CE_y%3D-k%5Cfrac%7Bq%7D%7Br_1%5E2%7Dsin%5Ctheta-k%5Cfrac%7Bq%7D%7Br_2%5E2%7D)
the second charge does not contribute for the x component of E.
To find r1 you use Pitagora's theorem:
![r_1=\sqrt{(0.150+0.150m)^2+(0.40m)^2}=0.500m](https://tex.z-dn.net/?f=r_1%3D%5Csqrt%7B%280.150%2B0.150m%29%5E2%2B%280.40m%29%5E2%7D%3D0.500m)
r2 = 0.40m
the angle is obtain by using a simple trigonometric relation:
![tan\theta=\frac{0.40}{0.150}=2.66\\\\\theta=tan^{-1}(2.66)=69.44\°](https://tex.z-dn.net/?f=tan%5Ctheta%3D%5Cfrac%7B0.40%7D%7B0.150%7D%3D2.66%5C%5C%5C%5C%5Ctheta%3Dtan%5E%7B-1%7D%282.66%29%3D69.44%5C%C2%B0)
Then, by replacing the values of r1, r1, q, theta and k you obtain:
![E_x=(8.98*10^9Nm^2/C^2)\frac{(6.00*10^{-9}C)}{(0.500m)^2}cos69.44=75.68N/C\\\\E_y=-(8.98*10^9Nm^2/C^2)(6.00*10^{-9}C)(\frac{sin69.44}{(0.500m)^2}+\frac{1}{(0.40m)^2})\\\\E_y=538.42N/C\\\\\vec{E}=75.68N/C\hat{i}-538.42N/C\hat{j}\\\\|\vec{E}|=\sqrt{(E_x)^2+(E_y)^2}=543.7N/C\\\\\theta=tan^{-1}(\frac{538.42}{75.68})=278\°](https://tex.z-dn.net/?f=E_x%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5Cfrac%7B%286.00%2A10%5E%7B-9%7DC%29%7D%7B%280.500m%29%5E2%7Dcos69.44%3D75.68N%2FC%5C%5C%5C%5CE_y%3D-%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%286.00%2A10%5E%7B-9%7DC%29%28%5Cfrac%7Bsin69.44%7D%7B%280.500m%29%5E2%7D%2B%5Cfrac%7B1%7D%7B%280.40m%29%5E2%7D%29%5C%5C%5C%5CE_y%3D538.42N%2FC%5C%5C%5C%5C%5Cvec%7BE%7D%3D75.68N%2FC%5Chat%7Bi%7D-538.42N%2FC%5Chat%7Bj%7D%5C%5C%5C%5C%7C%5Cvec%7BE%7D%7C%3D%5Csqrt%7B%28E_x%29%5E2%2B%28E_y%29%5E2%7D%3D543.7N%2FC%5C%5C%5C%5C%5Ctheta%3Dtan%5E%7B-1%7D%28%5Cfrac%7B538.42%7D%7B75.68%7D%29%3D278%5C%C2%B0)
hence, the magnitude of E is 543N/C with an angle of 278° from the positive x axis.