Question

Point P is located on the x-axis of a coordinate plane at x = 0.2 m.
Object A of charge -5e-9 C is located on the y-axis of a coordinate plane at y = 0.2 m.
Object B of charge 2e-9 C is located on the y-axis of a coordinate plane at y = -0.4 m.
Calculate the magnitude of the net electric field at point P due to objects A and B.

358.258 N ⁄ C
417.968 N ⁄ C
656.806 N ⁄ C
477.677 N ⁄ C
597.097 N ⁄ C

Point P is located on the x-axis of a coordinate plane at x = 0.1 m.
Object A of charge 1e-9 C is located on the y-axis of a coordinate plane at y = 0.2 m.
Object B of charge 3e-9 C is located on the y-axis of a coordinate plane at y = -0.5 m.
Calculate the direction (angle with respect to the positive x-axis) of the net electric field at point P due to objects A and B.

-27.096°
-25.996°
-28.196°
-30.396°
-29.296°

Answer 1

Both problems can be solved by using the following approach. First of all, the net electric field in the point P is equal to:

$$\vec{E} = \vec{E}_A+\vec{E}_B \\ \Rightarrow E_x=E_{Ax}+E_{Bx}, \, E_y = E_{Ay}+E_{By} \\ \Rightarrow E = \sqrt{E_x^2+E_y^2}, \gamma = \arctan{\frac{E_y}{E_x}}$$

We will also need the angles APO and BPO (assuming them to be positively defined)

$$\angle APO \equiv \alpha = \arctan{\frac{y_A}{x_P}}\\ \angle BPO \equiv \beta = \arctan{\frac{|y_B|}{x_P}}$$

and distances squared

$$AP^2\equiv r_A^2=x_P^2 +y_A^2\\ BP^2\equiv r_B^2=x_P^2 +y_B^2$$

Finally, the absolute values of the constituting electric fields and their projections:

$$E_A = k \frac{|q_A|}{r_A^2}, \, E_{Ax} = \pm E_A \cos{\alpha}, \, E_{Ay} = \pm E_A \sin{\alpha}\\ E_B = k \frac{|q_B|}{r_B^2}, \,E_{Bx} = \pm E_B \cos{\beta}, \, E_{By} = \pm E_B \sin{\beta}$$

The sign of the projections depends on the field vector direction.

Substituting the numerical values, we obtain:

1)

$$r_A^2 = 0.08, \, r_B^2 = 0.20\\ \alpha = 45^\circ, \, \beta = 63.435^\circ$$ $$E_A = 562.5 \, \frac{N}{C}, \, E_{Ax}=-397.748\, \frac{N}{C}, \, E_{Ay}=397.748 \, \frac{N}{C}\\ E_B = 90 \, \frac{N}{C}, \, E_{Bx}=40.249\, \frac{N}{C}, \, E_{By}=80.498 \, \frac{N}{C}$$

Finally,

$$E_x = -357.498 \, \frac{N}{C}, \, E_y=478.246 \, \frac{N}{C} \\ \Rightarrow E = 597.097 \, \frac{N}{C}$$

2)

$$r_A^2 = 0.05, \, r_B^2 = 0.26\\ \alpha = 63.435^\circ, \, \beta = 78.690^\circ$$

$$E_A = 180 \, \frac{N}{C}, \, E_{Ax}=80.498\, \frac{N}{C}, \, E_{Ay}=-160.997 \, \frac{N}{C}\\ E_B = 103.85\, \frac{N}{C}, \, E_{Bx}=20.366\, \frac{N}{C}, \, E_{By}=101.83 \, \frac{N}{C}$$

Finally,

$$E_x =100.864 \, \frac{N}{C}, \, E_y=-59.167 \, \frac{N}{C} \\ \Rightarrow \gamma = -30.396^\circ$$