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:
[tex]E_x=E_{1x}-E_{2x}\\\\E_x=k\frac{q}{r_1^2}-k\frac{q}{r_2^2}[/tex] ( 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):
[tex]E_x=0[/tex]
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:
[tex]\vec{E}=E_x\hat{i}[/tex]
[tex]E_x=k\frac{q}{r_1^2}+k\frac{q}{r_2^2}[/tex]
r1 = 0.300m+0.150m = 0.450m
r2 = 0.300m-0.150m = 0.150m
By replacing r1 and r2 in ( 1 ) you obtain:
[tex]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}[/tex]
c)
for P(0.150 , -0.40) you have both x and y components for E:
[tex]\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}[/tex]
the second charge does not contribute for the x component of E.
To find r1 you use Pitagora's theorem:
[tex]r_1=\sqrt{(0.150+0.150m)^2+(0.40m)^2}=0.500m[/tex]
r2 = 0.40m
the angle is obtain by using a simple trigonometric relation:
[tex]tan\theta=\frac{0.40}{0.150}=2.66\\\\\theta=tan^{-1}(2.66)=69.44\°[/tex]
Then, by replacing the values of r1, r1, q, theta and k you obtain:
[tex]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\°[/tex]
hence, the magnitude of E is 543N/C with an angle of 278° from the positive x axis.
