In a rectangular coordinate system a positive point charge q=6.00×10−9 C q=6.00×10−9 C is placed at the point x = +0.150 m, y =

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3 years ago

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In a rectangular coordinate system a positive point charge q=6.00×10−9 C q=6.00×10−9 C 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. Find the x- and y-components, the 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 = -0.400 m; (d) x = 0, y = 0.200 m.

Physics

1 answer:

jeka94 · Answer 1 · 2022-06-26T05:20:00+03:00

Answer:

a) E_net = 0

b) E_net = 2663.7 N/C

c) E_net = 525.915 N/C

d) E_net = 1380.864 N/C

Explanation:

Given:

- The positions of charges are = ( 0.15 , 0 ) & (-0.15 , 0 )

- Both charges have equal magnitude Q = +6*10^-19 C

Find:

Find the magnitude of the electric field at:

(a) the origin;

(b) x = 0.300 m, y = 0;

(c) x = 0.150 m, y = -0.400 m;

(d) x = 0, y = 0.200 m.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a) @ ( 0 , 0 )

- First we see that the charges +Q_1 and +Q_2 produce an electric field equal but opposite in nature because the distance r from origin and magnitude of charge is same. So the sum of Electric fields:

E_1 + E_3 = 0

Part b) @ ( 0.3 , 0 )

- The Electric field due to charge Q_1 is expressed by:

E_1 = k*Q_1 / r_1^2

Where the distance r from first charge to ( 0.3 , 0 ) is:

r_1 = 0.3 - 0.15 = 0.15 m

- The Electric field due to charge Q_2 is expressed by:

E_2 = k*Q_2 / r_2^2

Where the distance r from second charge to ( 0.3 , 0 ) is:

r_2 = 0.3 - (-0.15) = 0.45 m

- The net Electric Field at point ( 0.3 , 0 ) is:

E_net = E_1 + E_2

E_net = k*Q* ( ( 1 / r_1^2) + ( 1 / r_2^2) )

Plug values in:

E_net = (8.99*10^9)*6*10^-9* ( ( 1 / 0.15^2) + ( 1 / 0.45^2)))

E_net = 2663.7 N/C

Part c) @ ( 0.15 , -0.4 )

- The Electric field due to charge Q_1 is expressed by:

E_1 = k*Q_1 / r_1^2

Where the distance r from first charge to ( 0.15 , -0.4 ) is:

r_1 = 0.4 m

- The Electric field due to charge Q_2 is expressed by:

E_2 = k*Q_2 / r_2^2

Where the distance r from second charge to ( 0.15 , -0.4 ) is:

r_2 = sqrt (0.3^2 + 0.4^2) = 0.5 m

- The net Electric Field at point ( 0.15 , -0.4 ) is:

E_vertical = E_1 + E_2*sin(a)

E_horizontal = E_2*cos(a)

Where, a is the angle between x -axis and point ( 0.15 , -0.4 ):

cos(a) = 0.3 / r_2 = 0.3/0.5 = 3/5

sin(a) = 0.4 / r_2 = 0.4/0.5 = 4/5

Hence,

E_vertical = k*Q* ( ( 1 / r_1^2) + ( sin(a)/ r_2^2)

E_v = (8.99*10^9)*6*10^-9* ( ( 1 / 0.4^2) + ( 0.8 / 0.5^2 )

E_v = 509.733 N/C

And,

E_Horizontal = k*Q* ( ( 1 / r_1^2) + ( sin(a)/ r_2^2)

E_h = (8.99*10^9)*6*10^-9*( 0.6 / 0.5^2 )

E_h = 129.456 N/C

Hence,

E_net = sqrt ( E_vertical^2 + E_horizontal^2)

E_net = sqrt ( 509.733^2 + 129.456^2)

E_net = 525.915 N/C

Part d) @ ( 0 , 0.2 )

- The Electric field due to charge Q_1 is expressed by:

E_1 = k*Q_1 / r_1^2

Where the distance r from first charge to ( 0 , 0.2 ) is:

r_1 = sqrt (0.15^2 + 0.2^2) = 0.25 m

- The Electric field due to charge Q_2 is expressed by:

E_2 = k*Q_2 / r_2^2

Where the distance r from second charge to ( 0 , 0.2 ) is:

r_2 = sqrt (0.15^2 + 0.2^2) = 0.25 m

- The net Electric Field at point ( 0 , 0.2 ) is:

E_vertical = E_1*sin(a) + E_2*sin(a)

E_vertical = 2*E*sin(a)

E_horizontal = 0 ... Equal but opposite magnitude

Where, a is the angle between x -axis and point ( 0 , 0.2 ):

sin(a) = 0.2 / r_2 = 0.2/0.25 = 4/5

Hence,

E_vertical = 2*k*Q*sin(a)*( 1 / r_1^2)

E_v = 2*0.8*(8.99*10^9)*6*10^-9* ( 1 / 0.25^2)

E_v = 1380.864 N/C

Hence,

E_net = E_vertical

E_net = 1380.864 N/C