Electric potential/electric fields

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The discussion revolves around calculating the electric field and potential difference between two oppositely charged students when a spark jumps between their fingers at a distance of 7.0 cm. Participants express confusion about how to start the calculations, particularly regarding the force and charge involved. The electric field can be found using the formula E = F/q, while the potential difference can be calculated with V = kq/r. There is uncertainty about the value of charge (q) since it is not provided in the problem statement, leading to suggestions to assume a charge value typical for elementary particles. Clear guidance on how to approach the problem and the necessary formulas is sought by participants.
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Homework Statement



two oppositely charged students bring their fingers toward each other. a spark jumps between their fingers when brought 7.0 cm apart.

A) what is the electric field between fingers when sparking occurs

B) what is the potential difference between the two students

Homework Equations



q=fe
q=vc
f= kq1q2/r^2

The Attempt at a Solution



i think i can get part b if i get a
but i don't really have aclue where to even start
i know e=q/f
but what does distance have to do with it??
can someone help me where to start??
 
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i think you mean E = F/q...

use the force which has the distance (r) as the distance where the sparking occurs.

so F = (-kq^2)/(r^2), since they are oppositely charged particles...

I think the rest is self explanitory, you can solve for E using the force equation F = Eq, and your done.

for part b use the potential difference formula between the students. V = kq/r.
 
but what is q


can someone else please explain this
i don't undersatnd fubag
 
q should've been given, otherwise if you're assuming its an electron/proton type problem then use q =+/- 1.6 *10^-19 Coulombs...im not sure what the question is asking for
 
ya q isn't given at all
 
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