# Electric Field help

#### austin1250

1. Homework Statement
Alright i have two problems i need help with

1st.

Find the electric field at a point midway between two charges of +43.1 10^-9 C and +61.7 10^-9 C separated by a distance of 24.7 cm.

________ N/C

2nd.

A Van de Graaff generator is charged so that the magnitude of the electric field at its surface is 2.8 104 N/C.

(a) What is the magnitude of the electric force on a proton released at the surface of the generator?
________ N
(b) Find the proton's acceleration at this instant.
________ m/s^2

2. Homework Equations

E= K*q/r^2

3. The Attempt at a Solution
For the first one, I dont really know how to get the point in the middle. I got 9.1 x 10^3 N/C for the electric frield of the 61.7 c x 10^-9 and 6.4x10^3 N/C for the 43.1 c x10-9. So basicly what i need help with is the electric field at the mid-point.

For the 2nd one. I got part (a) right, i got 4.48 x10 ^-15 N. I need help getting the acceleration of the proton at that instant.

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#### LowlyPion

Homework Helper
For the first one, I dont really know how to get the point in the middle. I got 9.1 x 10^3 N/C for the electric frield of the 61.7 c x 10^-9 and 6.4x10^3 N/C for the 43.1 c x10-9. So basicly what i need help with is the electric field at the mid-point.

For the 2nd one. I got part (a) right, i got 4.48 x10 ^-15 N. I need help getting the acceleration of the proton at that instant.
(I'm not doing your math so if that's wrong you're still in trouble, but ...) Recall that the electric field is a vector field. At the point of interest you have calculated the individual scalar magnitudes of the vectors presumably at the midpoint. All you need to do now is determine the orientation of the two field contributions and add. By inspection they are both + charges so the field will necessarily be a difference of the 2, but directed away from the greater of the 2. (+ charges have outward directed fields.)

For 2)b) Doesn't F = m*a still?

For your first part add the two fields vectorially,they point in opposite directions.
For your second part use Newtons second law,you will need the mass of the proton.

#### austin1250

oh, ok i forgot about the mass of a proton, thanks, but im still stuck in the first question. i dont know if i did the math wrong but im not getting it right

#### LowlyPion

Homework Helper
oh, ok i forgot about the mass of a proton, thanks, but im still stuck in the first question. i dont know if i did the math wrong but im not getting it right

Distance is cm. Did you make it (.247/2) for your distance? That's a common oversight.

#### austin1250

Distance is cm. Did you make it (.247/2) for your distance? That's a common oversight.
Heres my calculations, and i did make it .247 before.

9x10^9 * 43.1 x10-9/.247^2 + 9x10^9 * 61.7 x10^-9/.247^2

is that correct?

comes out to be 6.4 x10^3 + 9.1 x 10^3 = 15500?

Then what? Do i divide by two?

#### LowlyPion

Homework Helper
Heres my calculations, and i did make it .247 before.

9x10^9 * 43.1 x10-9/.247^2 + 9x10^9 * 61.7 x10^-9/.247^2

is that correct?

comes out to be 6.4 x10^3 + 9.1 x 10^3 = 15500?

Then what? Do i divide by two?
No. Unfortunately you need to divide by 2 before you square your distances. So you could multiply your answer by 4 and that should undo not halving the denominator before you squared it.

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