# Ugh SO confused.

1. Jan 19, 2005

### DLxX

Could someone please take the time to do this problem and explain it to me so that I can do other similar problems on my own?

An electron moving at 1 percent the speed of light to the right enters a uniform electric field region where the field is known to be parallel to its direction of motion. If the electron is to be brought to rest in the space of 5.0cm, (a) what direction is required for the electric field, and (b) what is the strength of the field?

So far I've figured out what 1% of the speed of light is and converted 5cm into .05m. I then used the formula E = V/d to try and get the answer, but it was wrong. What am I doing wrong?

2. Jan 19, 2005

### stunner5000pt

Ok first of all we know for the electron it's initial velocity and final velocity anbd the distance to stop it in

so $$v_2 ^2 = v_1 ^2 + 2ad$$

now you found th acceleration.

Now acting on the electron is an electric field that should point from the left to right

Positive--------->electron goes like so -->>>>> Negative

the Force on the electron is F = Eq
and force = mass x accelerator

so $$F = m a = Eq$$

which gives $$E = \frac{ma}{q}$$

you should know the mass and charge for the electron which you can sub into that and solvef or E field

Last edited: Jan 20, 2005
3. Jan 20, 2005

### Staff: Mentor

It's impossible to say because you didn't give us enough information. Tell us exactly what you did: what numbers you substituted into which equations, and the results that you got. Then someone can probably tell you where you went wrong.

4. Jan 20, 2005

### dextercioby

Hold on a second:What's the electric potential V gotta do with this problem???I belive one of the previous posters presented you with a way to get to the solution...

Daniel.

5. Jan 20, 2005

### learningphysics

Another way to solve the problem is using conservation of energy. All the K.E is converted to electric potential energy.

(1/2)mv^2=qV

(1/2)mv^2=qEd

so solve for E in the above.

The previous poster made a mistake. Electric force = qE not E/q.

But that approach also yields the same answer.

6. Jan 20, 2005

### futb0l

I find this method easier, I think you should write down your methods, so we can then help you.