Physics, 06.05.2020 07:04 lulu8167

In a rectangular coordinate system a positive point charge q = 6.00 x 10^-9C is placed at the point x = +0.150 m, y = 0, and an identical point charge is placed at x = -0.150 m, y = 0. (a) Find the x- and y-components, magnitude and the direction of the electric field at the following points (a) the origin; (b) x = 0.300 m, y = 0; (c) x = 0.150 m, y = -0.40; (d) x = 0, y=0.200 m (a) 0 N/C (b) 2667 N/C (c) 526 N/C , 284 deg counter clockwise from positive x axis

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Answer from: noaj94

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.

for P(0,0), only there is the x component of the field because the point is the parallel line that connects both charges:

$E_x=E_{1x}-E_{2x}\\\\E_x=k\frac{q}{r_1^2}-k\frac{q}{r_2^2}$ ( 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

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:

$\vec{E}=E_x\hat{i}$

$E_x=k\frac{q}{r_1^2}+k\frac{q}{r_2^2}$

r1 = 0.300m+0.150m = 0.450m

r2 = 0.300m-0.150m = 0.150m

By replacing r1 and r2 in ( 1 ) you obtain:

$E_{x}=kq(\frac{1}{r_1^2}+\frac{1}{r_2^2})=(8.98*10^9Nm^2/C^2)(6.00*10^{-9})(\frac{1}{(0.450m)^2}+\frac{1}{(0.150m)^2})\\\\E_x=2660.7N/C\\\\\vec{E}=2660.7N/C\ \hat{i}$

for P(0.150 , -0.40) you have both x and y components for E:

$\vec{E}=E_x\hat{i}+E_y\hat{j}\\\\E_x=k\frac{q}{r_1^2}cos\theta+0N/C\\\\E_y=-k\frac{q}{r_1^2}sin\theta-k\frac{q}{r_2^2}$

the second charge does not contribute for the x component of E.

To find r1 you use Pitagora's theorem:

$r_1=\sqrt{(0.150+0.150m)^2+(0.40m)^2}=0.500m$

r2 = 0.40m

the angle is obtain by using a simple trigonometric relation:

$tan\theta=\frac{0.40}{0.150}=2.66\\\\\theta=tan^{-1}(2.66)=69.44\°$

Then, by replacing the values of r1, r1, q, theta and k you obtain:

$E_x=(8.98*10^9Nm^2/C^2)\frac{(6.00*10^{-9}C)}{(0.500m)^2}cos69.44=75.68N/C\\\\E_y=-(8.98*10^9Nm^2/C^2)(6.00*10^{-9}C)(\frac{sin69.44}{(0.500m)^2}+\frac{1}{(0.40m)^2})\\\\E_y=538.42N/C\\\\\vec{E}=75.68N/C\hat{i}-538.42N/C\hat{j}\\\\|\vec{E}|=\sqrt{(E_x)^2+(E_y)^2}=543.7N/C\\\\\theta=tan^{-1}(\frac{538.42}{75.68})=278\°$

hence, the magnitude of E is 543N/C with an angle of 278° from the positive x axis.

Answer from: Quest

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Answer from: Quest

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