Finding particle speed in electric field and zero points between charges

[cnugrl]

I'm not sure if this has been asked, but i am in major need of help. i have 2 questions. first is determining the speed of a particle when the speed is parallel to the electric field. i talked to my prof. and he suggested finding the energy, U, and by integrating E with respect to z, by doing so, that left me with E*z from Zinitial (1-cos(theta)*L) to Zfinal(0). after doing this, i st this equal to 1/2 mv^2. to solve for the velocity, however, i am not getting the correct answer. and I'm getting frustrated. does anyone know where i am going wrong?

my second question is finding finite value(s) where an electric field is zero. by having a +q charge at the orgin and a -2q charge at x=5.8m. my professor told me to use kq/r^2 and where q is and q or -2q and r is x and x-d respectively where d is 5.8 m. i solving and put the equation = 0, and used the quadratic formula to get the answer. However, i am not getting this correct, when i solve for x i get x^2+2xd-d^2. i have used various other ways of trying to get this answer correct, by changing the signs and and i am still getting incorrect answers. I need to do the same with electric potential, but i do not understand how i can get 2 answers from a linear equation. i am getting d-3x where d =5.8 m.
if anyone could help, i would appreciate it so much

 

[dextercioby]

I'm not sure if this has been asked, but i am in major need of help. i have 2 questions. first is determining the speed of a particle when the speed is parallel to the electric field. i talked to my prof. and he suggested finding the energy, U, and by integrating E with respect to z, by doing so, that left me with E*z from Zinitial (1-cos(theta)*L) to Zfinal(0). after doing this, i st this equal to 1/2 mv^2. to solve for the velocity, however, i am not getting the correct answer. and I'm getting frustrated. does anyone know where i am going wrong?

Who's "Z"...?

The equation is:
$$m\frac{d\vec{v}}{dt}=q\vec{E}$$

If the acceleration has the sense with the velocity (the sense of the acc.is the same with the one of the electric field for a positively cherged particle),then project the previous relation on an axis and integrate wrt to corresponding limits...

my second question is finding finite value(s) where an electric field is zero. by having a +q charge at the orgin and a -2q charge at x=5.8m. my professor told me to use kq/r^2 and where q is and q or -2q and r is x and x-d respectively where d is 5.8 m. i solving and put the equation = 0, and used the quadratic formula to get the answer. However, i am not getting this correct, when i solve for x i get x^2+2xd-d^2. i have used various other ways of trying to get this answer correct, by changing the signs and and i am still getting incorrect answers. I need to do the same with electric potential, but i do not understand how i can get 2 answers from a linear equation. i am getting d-3x where d =5.8 m.
if anyone could help, i would appreciate it so much

No,on normal basis it should reduce to a quadratic...

Daniel.

 

[cnugrl]

Z is the limit bounds...initial to final. plus, i was using it as a variable to integrate on