Question

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.

Answer 1

Answer:

  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 = $k \frac{q}{r^2}$

charge 1

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

                 r₁ = $\sqrt{ (x_1-x_o)^2 + (y_1-y_o)^2}$

                 r₁ = $\sqrt{0 +(1.04-0)^2}$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