How Does Coulomb's Law Apply to Electric Fields at Points A and B?

[Shakas]

Homework Statement

Use Coulomb's law to determine the magnitude and direction of the electric field at points A and B in Fig. 16-57 due to the two positive charges (Q = 8.0 µC) shown. Is your result consistent with Fig. 16-31b?

Figure 16-57

Figure 16-31b

A)Find the magnitude of the electric field at point A and direction.
B)Find the magnitude of the electric field at point B and direction.

Homework Equations

E=kQ/r^2

The Attempt at a Solution

A)
Let the left Q charge be Q1
Let the right Q charge be Q2

I started by finding the distance vector between Q1 and A, which is the same for Q2 and A.

sqrt((0.1^2)+(0.05^2))=.111m

I then found the electrical field between Q1 and A, which is the same for Q2 and A, but in the opposite direction.

E=(9e9)(8e-6)/(.111^2)=5.84e6

I then split the electrical field into x and y components. Used law of sines to find the angle.

sin(90)/(.111)=sin(x)/(.05)

x=26.8 degrees

Splitting into x and y components:

Q1 to A.
5.84e6*sin(27)=2.651e6 up
5.84e6*cos(27)=5.203e6 right

Q2 to A.
5.84e6*sin(27)= up
5.84e6*cos(27)=5.203e6 left

The x-components cancel due to opposite directions and the y-components are combined.

2.651e6 + 2.651e6 = 5.3e6 N/C 90 degrees above the horizontal.

This is apparently the wrong answer and I do not know what I am doing wrong...

B)
I'm not going to attempt B until I get part A right...

 

[mjsd]

interesting, I don't see a problem with your workings either

 

[ogb p]

Dear student,

Your attempt at solving this problem is a good start, but there are a few errors in your calculations. First, the distance between the charges and point A should be 0.111m, not 0.11m. Additionally, when you calculated the electrical field between Q1 and A, you should have used the distance of 0.111m, not 0.05m. This is because the electrical field is a vector quantity and its direction depends on the distance between the charges and the point of interest.

Also, when you split the electrical field into x and y components, you should have used the angle of 63.2 degrees (90-26.8) for both Q1 and Q2. This is because the angle between the x-axis and the direction of the electrical field is 90 degrees minus the angle you calculated. Using the correct angle, the x-component for Q1 should be 2.651e6*cos(63.2) = 1.203e6 to the right, and for Q2 it should be 5.203e6*cos(63.2) = 2.358e6 to the left. The y-components should be the same as what you calculated, 2.651e6 up for both Q1 and Q2.

After combining the x-components, you will get a net electrical field of 1.203e6 to the right, which is consistent with the direction shown in Figure 16-31b. Similarly, at point B, the net electrical field will be to the left, also consistent with the direction shown in Figure 16-31b.

I hope this helps clarify your confusion and guide you towards the correct solution. Keep up the good work!