A point charge of -4 mu or micro CC is located at x = 4 m, y = -2 m. A second point charge of 12 mu or micro CC is located at x = 1 m, y = 4 m.

(a) Find the magnitude and direction of the electric field at x = -1 m, y = 0.


I got 4545.69 N/C by using this equation E= kQ/r^2 (vector r) and finding EQ1 and EQ2 then finding their x and y components and adding those then taking the magnitude. but my answer is wrong.

1 answer

You are on the right track; however, you may have made a mistake with calculating the distances or summing the components. Let's do the calculation again.

(a) To find the electric field at point P(-1, 0), we need to find the fields due to each charge separately and then sum the contributions.

Let's denote the point where the -4 μC charge is located as A(4, -2) and the point where the 12 μC charge is located as B(1, 4).

The distance vector from A to P is r1 = <-5, 2> and from B to P r2 = <-2, -4>.

The magnitudes of the distance vectors are |r1| = sqrt((-5)^2 + 2^2) = sqrt(29) and |r2| = sqrt((-2)^2 + (-4)^2) = sqrt(20).

Now, let's find the electric fields due to each charge at point P:

E1 = k * q1 / |r1|^2 * r1 = (9 × 10^9 N m^2/C^2) * (-4 × 10^-6C) / (29 m^2) * <-5, 2>
E1 = <-60, 24> N/C

E2 = k * q2 / |r2|^2 * r2 = (9 × 10^9 N m^2/C^2) * (12 × 10^-6C) / (20 m^2) * <-2, -4>
E2 = <108, 216> N/C

Now, let's find the total electric field at point P by summing the components of E1 and E2:

E_total = E1 + E2 = (<-60 + 108, 24 + 216>) = (48, 240) N/C

The magnitude and direction of the total electric field are:

|E_total| = sqrt(48^2 + 240^2) ≈ 246.6 N/C

tan(theta) = 240/48 => theta ≈ 78.7° (above the negative x-axis in the second quadrant)

Therefore, the magnitude and direction of the electric field at point P are approximately 246.6 N/C and 78.7° above the negative x-axis.