Electric fields and parallel plates

[F.B]

I need help on a few questions.

1.An alpha particle has a positive charge of 2e and a mass of 6.6 x 10^-27 kg. With what velocity would the particle reach the negative plate of a parallel plate apparatus with a potential difference of 2.0 x 10^3 V.
a)if it started from rest at the positive plate.
b)if it started from rest at a point halfway between the plates.

I got the answer for a, it 4.4 x 10^5 m/s. But i don't know how to do b.

2. An electron is accelerated through a uniform electric field of magnitude 2.5 x 10^2 N/C with an initial speed of 1.2 x 10^6 m/s parallel to the electric field.
a)Calculate the speed of the electron after it has traveled 2.5 cm in the field.

I don't know how to do this one at all.

3. Two electrons are fired at 3.5 x 10^6 m/s directly at each other.
a)Calculate the smallest possible between the two electrons.

I tried do Ee=Ek but that didnt work i kept getting the wrong answer. So can anyone please help me.

 

[Astronuc]

Well how did you do part 'a' of question 1? In part b, the acceleration takes place over half the distance. Hint: The electric field is contant between paralle plates.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefie.html

2. What is the force on a charge in an electrical field? Get F, and then solve for a. Then what is the equation for final velocity, initial velocity, acceleration and distance.

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot1

3. Each electron has a kinetic energy and approach each other until they stop at some distance, at which point the energy is all electrical potential energy.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html

 

[quicknote]

1. For part b, we can use the equation for acceleration in an electric field: a = qE/m, where q is the charge of the particle, E is the electric field strength, and m is the mass of the particle. Since the particle is starting from rest at a point halfway between the plates, it will have traveled half the distance between the plates (d/2) before reaching the negative plate. Therefore, we can use the formula for displacement in uniform acceleration: d = 0.5at^2, where a is the acceleration and t is the time. We can solve for t and then use it to find the final velocity using the equation v = u + at, where u is the initial velocity (0 in this case). The final velocity will be the same as the velocity in part a, since it is the same particle and the potential difference is the same. So, the final velocity for part b will also be 4.4 x 10^5 m/s.

2. To calculate the speed of the electron after traveling 2.5 cm in the field, we can use the formula for displacement in uniform acceleration: d = ut + 0.5at^2, where u is the initial velocity, a is the acceleration, and t is the time. In this case, u = 1.2 x 10^6 m/s, a = 2.5 x 10^2 N/C, and d = 2.5 cm = 0.025 m. We can solve for t and then use it to find the final velocity using the equation v = u + at. The final velocity will be the same as the initial velocity, since the electron is traveling parallel to the electric field and there is no change in its velocity in that direction. So, the final velocity will also be 1.2 x 10^6 m/s.

3. To calculate the smallest possible distance between the two electrons, we can use the equation for the electric force between two charged particles: F = kq1q2/r^2, where k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them. Since the two electrons have the same charge (q1 = q2 = -e), the force between them will be repulsive. We can equate this force to the kinetic energy of the electrons using the