# Find electric field strength between disks

1. Feb 12, 2013

### johnnyboy53

1. The problem statement, all variables and given/known data

Two 2.0cm diameter disks face each other, 1.0mm apart. They are charged to -10nC and
10nC.

-What is the field strength between the disks?

-A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?

2. Relevant equations
Q = (10*10^-9)
ε0=(8.85*10^-12)
A = area of circle = πr^2
E = E(+) + E(-) = Q/ε0A

vf^2 = vi^2 +2Ax

3. The attempt at a solution

(10*10^-9)/(8.85^10^-12)(π)(.1^2) = 35.5 but the book gives it as (3.6 * 10^6) N/C

b: vf^2 = vi^2 + 2AX
vi= sqrt( vf^2+2(3.6*10^6)(.001))
= 84.85
book has it as 8.5 * 10^5

I think i am supposed to convert the distances meters or I need to do some type of conversion to get from the original answer to N/C units. I get the right answer but the precision is off. What could i be missing? Thanks!

2. Feb 12, 2013

### Staff: Mentor

Check the value you've used for the radius; looks like a conversion error. The result lacks units and the powers of ten.
The formula is incomplete; You've used the field strength in place of the acceleration which would not yield the correct units for the results. Determine the acceleration A that the proton will experience and then plug that into your formula.

Once again you've dropped the powers of ten and the units, making the result incorrect regardless of the numerical value.

Last edited: Feb 12, 2013
3. Feb 12, 2013

### johnnyboy53

oh 1 cm = .01m. what do you mean by dropping powers of 10?

Instead of writing .01m, do i keep it as (1 * 10^-2)?

4. Feb 12, 2013

### rude man

I would use energy conservation to determine vi. BTW vf = 0.

5. Feb 12, 2013

### Staff: Mentor

Yup.
The field won't be as large as 35.5 N/C. There will be a power of ten multiplying it, which you've dropped.
Either way is fine so long as you retain the order of magnitude in the calculations.