The electric force by the surface of the earth, help please

  • Thread starter Simon316
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  • #1
I'm having a little problem with this one

The electric field of the earth (surface), is about 120 V/m.

1. what is the force on a oxygen-ion in this field?

Okey we have 120 V/m = 120 N/C,
so E = 1/(4π€0) * q/(r^2) = 120 N/C (electric fiels equation), where 1/(4π€0) = 9*10^9N*m^2/C^2.

An oxygen has 2 excess electrons, so it’s charge must be 1,602*10^-19C * 2 = 3,204*10^-19 = q
We can isolate r to be r =√(q/(E/(9*10^9))) = 4,902*10^-6m= 0,005mm (this seems to small I think)

Now I would use coulombs law to find the force F = 1/(4π€0) * q1q2/(r^2), but I don’t have q2 .

I don’t know if I’m on the right track here, any help would be highly appreciated, thx.

2. find the acceleration of this ion, treating it as if it had no contact with other molecules or ions.

Should I use newton's second law here?

Kind regards


Answers and Replies

  • #2
Science Advisor
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Treat this as the force on a charge in a uniform E field. F=E*q. Forget the r^2 bit. Are you sure they mean doubly ionized oxygen?
  • #3
oh yea sorry I left that out, it says dobbelcharged oxygen molecule.

okey F=E*q makes sense, a bit easier to:)

thanks alot