How do I calculate the force on a charged particle in an electric field?

[ghostbuster25]

I have a homework question with regards to a charged particle in an electric field.
The particle has a charge of -4e
The field is constant at 45000 N C^1



For this i have been using the equation F_el= q E(r)



The trouble i have been having is trying to work out -4e in Coloumbs.
I know -e=1.602 * 10^-19 coloumbs but is it as simple as 4 * 1.602 * 10^-19?
It just seems to give me a really tiny force in Newtons. I know its only a small particle but something tells me its just not right.

Any help if I am going the right or wrong way would be much appreciated :)

 

[RoyalCat]

Yep. The charge of an electron is just that, a unit of charge. If you have 4 times the charge of one electron, you have 4*e of charge.

The force is small, since the charge is very small.
To put it in proportion, let's look at the mass of the electron and the acceleration it experiences as a result of the electric field. You should note, that the field we're dealing with here is MASSIVE in terms of laboratory fields.

Writing out Newton's Second Law for a stationary electron (Ignoring the directions of force, acceleration and field):

F=ma

F=qE

qE=ma

a=\frac{q}{m}E

m=9.1\cdot 10^{-31} kg

q=1.6\cdot 10^{-19} C

E=45,000 \frac{N}{C}

a=7.9\cdot 10^{15} \frac{m}{s^2}

That force doesn't seem too tiny now, does it?

 

[ghostbuster25]

Thanks for your quick reply. I agree it is a massive force and am a bit worried about my previous calculations now.

The field is produced by a fixed charge of 2.00*10^-11C at a diastance of 2mm

I used the equation for the field
E(r) = Q/ 4pi*E0*r^2

which is electric field at distance r is equal to point charge Q divded by 4pi * the value for the quantity of the permittivity of free space (9*10^9) *( 2.00*10^-3)^2

I arranged to equal 9*10^9 * Q/r^2

That gave me 45000 N C^-1

I hope that makes sense :)