Point charge - magnitude calculations

[swimmer052005]

Homework Statement

A 4.9-µC point charge is located at x = 1.1 m, y = 3.1 m, and a -3.9-µC point charge is located at x = 2.1 m, y = -1.9 m.

(a) Find the magnitude(inkN/C) and direction(in degrees) of the electric field at x = -3.2 m, y = 0.9 m.

(b) Find the magnitude(in N) and direction(in degrees) of the force on a proton at x = -3.2 m, y = 0.9 m

Homework Equations

The Attempt at a Solution

We haven't been taught this in class specifically, so i donn't know how to go about it. Help please!

 

[espen180]

Use the equation for the Coloumb force:

$$\vec{F}=k_e \frac{q_1q_2}{r^2} \hat{r}$$

And the electric field

$$\vec{E}=\frac{\vec{F}}{q}=k_e\frac{q}{r^2}$$

and the superposition principle (E=E₁+E₂)

 

[kerimek]

As a scientist, it is important to first understand the basic principles and equations related to point charges and electric fields. The electric field at a certain point in space is given by the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the magnitude of the point charge, and r is the distance between the point charge and the point in space. The direction of the electric field is given by the direction of the force that a positive test charge would experience at that point.

(a) To find the magnitude of the electric field at x = -3.2 m, y = 0.9 m, we can use the equation mentioned above. First, we need to find the distance between the point charges and the point in space. Using the distance formula, we get: r1 = √[(1.1 - (-3.2))^2 + (3.1 - 0.9)^2] = √(16.5^2 + 2.2^2) = √273.29 ≈ 16.53 m and r2 = √[(2.1 - (-3.2))^2 + (-1.9 - 0.9)^2] = √(5.3^2 + (-2.8)^2) = √34.57 ≈ 5.88 m.

Next, we can calculate the electric field due to each point charge using the equation E = kQ/r^2. For the 4.9-µC point charge, we get E1 = (9x10^9 Nm^2/C^2)(4.9x10^-6 C)/(16.53 m)^2 ≈ 1.01x10^4 N/C. For the -3.9-µC point charge, we get E2 = (9x10^9 Nm^2/C^2)(-3.9x10^-6 C)/(5.88 m)^2 ≈ -1.31x10^4 N/C.

To find the net electric field at the given point, we can use vector addition and get E = E1 + E2 = (1.01x10^4 N/C) + (-1.31x10^4 N/C) = -0.3x10^4 N/C. The magnitude of the electric field is therefore