# Magnitude and direction of electric force (file attached)

1. Jul 28, 2007

### trah22

1. The problem statement, all variables and given/known data
I was kind of confused on how the radius for the charge on q3 by q1 is 2a^2, i attached the file for my notes regarding this. The question is at the bottom of the pdf scan.

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### physics5.PDF
File size:
31.9 KB
Views:
51
Last edited: Jul 28, 2007
2. Jul 28, 2007

### Staff: Mentor

The needed distance is the hypotenuse of the right triangle whose sides are both "a". Hint: Pythagorean theorem.

3. Jul 28, 2007

### trah22

c^2=a^2+b^2

so

c^2=a^2+a^2
c^2=2a^2

what about the ^2 for the c, is it just left there? because it seems like it since r=2a^2

4. Jul 28, 2007

### Staff: Mentor

Careful: it's r^2 = 2a^2, not r = 2a^2.

5. Jul 28, 2007

### trah22

On this other page of notes which has to do with the introduction of acceleration in electric fields, there are three parts of the problem that im not quite following, i figured it would be easier if i just wrote it down by hand on the actual note sheet than typing it all out.

For the cosign question, i dont undestand what happened to it because the sin is still there in the next step but not the cos.

#### Attached Files:

• ###### physics6.PDF
File size:
41.5 KB
Views:
46
Last edited: Jul 28, 2007
6. Jul 28, 2007

### trah22

If what im asking isnt really quite clear, i can write it out again more specifically and scan it.

7. Jul 29, 2007

### Staff: Mentor

(Be sure to include the full problem statement next time.)

(1) Why negative? The field points down (thus is negative), the charge is positive: so the force and acceleration are both downward, thus negative.

(2) They are solving for the value of time that makes y = 0. Set that factor equal to zero and solve for t.

(3) Regarding what happened to the cosine: note that $\theta$ changed to $2\theta$. (Review your trig identities for $\sin\theta \cos\theta$.)