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.

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

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.