Two test charges are located in the x–y plane. If q1=−2.600 nC and is located at x1=0.00 m, y1=1.0400 m, and the second test charge has magnitude of q2=3.600 nC and is located at x2=1.400 m, y2=0.400 m, calculate the Ex and Ey components, of the electric field ⃗ in component form at the origin, (0,0). The Coulomb force constant is 1/(40)=8.99×109 N·m2/ C2.

  Eₓ = -4,187 N / C,       E_y = 6,937 N / C

Explanation:

To solve this exercise we will calculate the electric field at the desired point (0, 0) and then add it vectorially

                 E =

charge 1

the value of q₁ = - 2,600 nC = -2,600 10⁻⁹ C

                 r₁ =

                 r₁ = RA 0 + (1.04 - 0) 2

                 r₁ = 1.04 m

we substitute

                 E₁ = k q1 / r12

                 E₁ = 9 10⁹   2.6 10⁻⁹ / 1.04²

                 E₁ = 21.635 N / C

This electric field is directed towards the negative charge and is in the direction of the y axis.

                 E₁ = 21.635 j ^   N / C

charge 2

the value of q₂ = 3,600 nC = 3,600 10-9 C

                 r₂ = Ra (1.4 -0) 2 + (0.4 -0) 2

                 r₂ = 1.4560 m

we substitute

                 E₂ = k q₂ / r₂²

                 E₂ = 9 10⁹ 3,600 10⁻⁹ / 1.4560²

                  E₂ = 15.283 N / C

This field leaves the charge since the charge is positive, let's find the angle

               tan θ = y / x

               θ = tan⁻¹ y / x

               θ = tan⁻¹ 0.400 / 1.400

               θ = 15.9º

let's decompose the electric field

               sin 15.9 = E_{2y} / E2

               cos 15.9 = E₂ₓ / E2

               E_{2y} = E2 sin 15.9

               E₂ₓ = E2 cos 15.9

                E_{2y} = 15.283 sin 15.9 = 4.187 N / C

               E₂ₓ = 15.283 cos 15.9 = 14.698 N / C

the field lines leave the positive charge and are directed to the left, therefore

               E_{2y} = - 4.187 N / C

               E₂ₙ = -14.698N / C

the total field on each axis is

               Eₓ = E_{1x} + E_{2x}

               Eₓ = 0 -4,187

                Eₓ = -4,187 N / C

                E_y = 21.635 - 14.698

                 E_y = 6,937 N / C

f is prop to m1xm2 double m1 double force

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this