# Homework Help: Electric Forces and Electric Fields of a proton

1. Sep 13, 2006

### himura137

Hi, I am currently stuck on the following questions. Any help is greatly appreciated. thank you
*all question are algebra based

1) A point charge of -0.90 microC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A point charge q is fixed to each of the remaining corners. The net force acting on either of the charges q is zero. Find the magnitude of q.
- i keep getting 3.6 micro C

2) A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and increases its linear momentum to 8.6 x 10^-23 kg·m/s from 2.4 x 10^-23 kg·m/s in a time of 5.7 x 10^-6 s. What is the magnitude of the electric field?

3) A uniform electric field has a magnitude of 3.1 x 10^3 N/C. In a vacuum, a proton begins with a speed of 2.5 x 10^4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 1.5 mm.

2. Sep 13, 2006

### andrevdh

1) How did you get to $3.6\mu C$?

3. Sep 13, 2006

### himura137

i put one of the q as q1 and the -0.9uc as q2 and q3 and the other q as q4 and used the equation F= (k|q1||q2| / r2) +(k|q1||q3| / r2) + (k|q1||q4| / r2) and since the F = 0 on the q and all of them contain k and q1, i just divide it straight through and got ( |q2| + |q3| ) 2 = q which when i plug in the 0.9uc for them, i get 3.6 uc

4. Sep 13, 2006

### Staff: Mentor

But q4 is not r2 away from the target q.

EDIT -- And the forces on q from q1 and q2 are not pointing in the same directions as the force from q4 on q. You need to do a little trig to get the correct answer for the magnitude of q4.

Last edited: Sep 13, 2006
5. Sep 13, 2006

### himura137

i placed the charged like this
q2 q4
q1 q3
i split the question into x and y and got
Fx = (k|q1x||q3x|/ 1^2 ) - (k|q1x||q4x|/ (*root*2)^2 )(cos 45)
Fy = (k|q1y||q2y|/ 1^2 ) - (k|q1y||q4y|/ (*root*2)^2 )(sin 45)
and i get
Fx and Fy as 2.54558 and that is the answer but not why its 2.54558, i though I am supposed to do something else after i got that number since its only the force due in one direction

can someone help me with the other 2 problems, thank you

Last edited: Sep 13, 2006
6. Sep 13, 2006

### himura137

#2
i used the V= Vo +at to find the acceleration of the proton, then multiply that by the mass of proton (1.67262 x 10^-27 kg) to get the Force, than used the equation E = F/q where q is the charge on an electron and since its a proton its positive with the unit (1.6022×10-19 C) and i get 1.355268 x 10^-25

#3, i used the equation F = Eq = ma to find the acceleration of the proton, then used it in the equation V^2 = Vo^2 + 2ad and i got 25017.8105

7. Sep 14, 2006

### himura137

*bumped* i showed my work, can someone please check

8. Sep 14, 2006

### andrevdh

Let's do #1 another way. Consider one ot the q charges. It is experiencing three forces two of $Fqq_1$ and one of $Fqq$. The resultant of the two perpendicular components of $Fqq_1$ need to balance that of $Fqq$ for equilibrium to excist. So we can form a perpendicular triangle with these three mentioned forces.

9. Sep 14, 2006

### andrevdh

You have all the components of all of the forces on q1. Which means that Fx and Fy should both be zero! So I do not understand why you say that both components are 2.54558.

#2 I agree with your answer for this problem. So the given answer is most likely wrong.

#3 I got another answer for its final speed. What was your acceleration?

Last edited: Sep 14, 2006
10. Sep 14, 2006

### himura137

Fx= 0 = (k|q1x||q2x|/ 1^2 ) + (k|q1x||q3x|/ 1^2 ) - (k|q1x||q4x|/ (*root*2)^2 )(cos 45)

Fy= 0 = (k|q1y||q2y|/ 1^2 ) + (k|q1y||q3y|/ 1^2 ) - (k|q1y||q4y|/ (*root*2)^2 )(sin 45)
so i actually found the force of q4x and q4y, which is same as q1x and q1y

can you check #2 and #3 for me also

#2
i used the V= Vo +at to find the acceleration of the proton, then multiply that by the mass of proton (1.67262 x 10^-27 kg) to get the Force, than used the equation E = F/q where q is the charge on an electron and since its a proton its positive with the unit (1.6022×10-19 C) and i get 1.355268 x 10^-25

#3, i used the equation F = Eq = ma to find the acceleration of the proton, then used it in the equation V^2 = Vo^2 + 2ad and i got 25017.8105