Calculating Electric Force on an Electron Near a Point Charge

[dois]

Homework Statement

You are observing the effects of single point charge with a magnitude of +4.3muC.
If a singe electron is placed 10cm to the right of this charge, what is the magnitude and direction of the electric force it will experience?

Homework Equations

E= Fe/q
E= kq/r^2

The Attempt at a Solution

E=kq/r^2

r=10cm= 0.10m
k=9.0x10^9
q=4.3muC= 4.3x10^-6

(9.0x10^9)(4.3x10^-6)/(0.10)^2
=3.87x10^6

I'm not sure if this is right, or if I'm even using the right equation.

 

[ojs]

You are close.

If you look at the first equation and just think about the dimensions, the dimension of an electric field is V/m or N/C, that is not the dimension of the equation E = Fe/q (unless the e there is not supposed to be the electric charge of an electron).

Also, the second equation is not quite right either, one little thing missing, can you find what?

 

[dois]

Could we use,
E=kq1q2/r^2 ?

 

[ojs]

yes, exactly, that is the correct equation to find an electric field between two point charges.

Now how do we use that to find the force between those two charges?

 

[dois]

E=kq1q2/r^2
k= 9.0x10^9
q1=4.3x10^-6
q2= 1e= 1.602x10^-19
r=0.10
=(9.0x10^9)(4.3x10^-6)(1.602x10^-19)/(0.10)^2
=6.19x10^-13

 

[ojs]

Sorry, I mixed up equations there. The equation that you give here and the value of 6.19x10^-13N is the force between the charges, your original equation for the electric field E = kq/r^2 is correct.

You can also look at it from the point of view that you have a 4.3 microC charge which produces an electric field, then an electron is put in that electric field and feels a force that is equal to Ee (where E is the electric field from the 4.3 microC charge and e is the charge of the electron) which gives you the equation

F = kq1q2/r^2

Sorry about the mixup, hope you understand it nonetheless.

 

[dois]

So what I originally did is correct?

 

[cupid.callin]

So what I originally did is correct?

Yes you were correct

$$F = \frac{kq_1q_2}{r^2}$$
$$E = \frac{kq}{r^2}$$

 

[ojs]

Not quite right, almost.

You had the right equation for the electric field produced by the 4.3 microC charge but you didn't calculate the force the electron felt in that field which is 3.87x10^6N/C x 1.602x10^-19C = 6.19x10^-13N

 

[cupid.callin]

Not quite right, almost.

You had the right equation for the electric field produced by the 4.3 microC charge but you didn't calculate the force the electron felt in that field which is 3.87x10^6N/C x 1.602x10^-19C = 6.19x10^-13N

Oh yes!

I forgot to mention that after reading the latter confusing posts.

 

[saiyaex]

the direction would be to the left towards the charge, right?