Calculating the magnitude of point charges

[ezperkins]

Homework Statement

A point charge of -6 µC is located at x = 1 m, y = -2 m. A second point charge of 12 µC is located at x = 1 m, y = 3 m.

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


(b) Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0.

Homework Equations

The Attempt at a Solution

$$\Sigma_{F}= k\frac{-6 \mu C}{8} + k\frac{12 \mu C}{13} = 1.08e4 \frac{N}{C}$$


That's for the magnitude part of A. I've already submitted to many incorrect responses, but I would like to know how to do it. Thanks ahead of time.

 

[hage567]

Homework Statement

A point charge of -6 µC is located at x = 1 m, y = -2 m. A second point charge of 12 µC is located at x = 1 m, y = 3 m.

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


(b) Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0.

Homework Equations

The Attempt at a Solution

$$\Sigma_{F}= k\frac{-6 \mu C}{8} + k\frac{12 \mu C}{13} = 1.08e4 \frac{N}{C}$$


That's for the magnitude part of A. I've already submitted to many incorrect responses, but I would like to know how to do it. Thanks ahead of time.

You need to find both the x and y components from each charge at the point of interest and add them up accordingly to find the resultant field.

 

[ezperkins]

I did that for another problem that asked for the respective components, why is it necessary that I so that for this problem?

 

[hage567]

Since I don't know what the other question was, I can't tell you what's different. You need to do vector components because the point of interest is not on the same line as the two charges. It is a two dimensional situation. Have you tried what I suggested?

 

[rwisz]

I would like to offer some guidance on how to correctly approach this problem. First, it is important to understand the concept of electric fields and forces. Electric fields are created by charges and can exert a force on other charges. The magnitude of an electric field at a point is defined as the force per unit charge at that point. In this case, we have two point charges, one with a negative charge of -6 µC and one with a positive charge of 12 µC.

To calculate the magnitude of the electric field at a specific point, we can use the formula:

E = kq/r^2

where E is the electric field, k is the Coulomb's constant (9x10^9 Nm^2/C^2), q is the charge, and r is the distance from the point charge to the point where we want to calculate the field.

(a) To find the magnitude of the electric field at x = -1 m, y = 0, we need to first find the distance from this point to each of the point charges. For the first point charge at x = 1 m, y = -2 m, the distance is 3 m (using the Pythagorean theorem). For the second point charge at x = 1 m, y = 3 m, the distance is also 3 m. Now, we can plug in these values into the formula:

E = kq/r^2

For the first point charge, the electric field will be:

E1 = (9x10^9 Nm^2/C^2)(-6x10^-6 C)/(3 m)^2 = -6x10^3 N/C

Note that the negative sign indicates that the electric field is directed towards the point charge.

For the second point charge, the electric field will be:

E2 = (9x10^9 Nm^2/C^2)(12x10^-6 C)/(3 m)^2 = 12x10^3 N/C

Adding these two electric fields together (since they are in opposite directions), we get:

E = E1 + E2 = -6x10^3 N/C + 12x10^3 N/C = 6x10^3 N/C

Therefore, the magnitude of the electric field at x = -1 m, y = 0 is 6x10^3 N/C, directed