Electric Forces and Electric Fields

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.

andrevdh

1) How did you get to [itex]3.6\mu C[/itex]?

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

berkeman

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.

himura137

ok i got the answer
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

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

both answer are wrong, not sure where i messed up on, please help

himura137

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

andrevdh

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

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?

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