# Question on electrostatics

1. Jan 31, 2006

### SS2006

the formula

v = kq1/r +kq2/r

if a charge is -2 micro couloumbs for instance is that put into the equatoin and it becomes minused instead or the sign doesnt effect q in the formula

cuase i remmeber in the f = kq1q2
the sign has no effect

2. Jan 31, 2006

### DaMastaofFisix

Mkay, for your first concern, the formula for voltage is correct. The thing is though is that the formula has q1 and q2, which are different charges, which can either be positive or negative. And yes, if the charge is negative, then yes, that particular term of the equation (which by the way, you can use for as many charges as you'd like) becomes negative.

AS for your second claim, the electric force between two particles is kq1q2/r^2! Very important is the sign(how much like yoda did I just sound there). In this case, the charge of the q's will determine whether the force will be attractive (negative sign) or repulsive (positive). Hopefully that asnwered your question.

3. Jan 31, 2006

### SS2006

ur right about the force part, i meant it i dont use it in the equation to find the Force magnitude, then i can draw FBD's to see which way itll go , cause of positive or negative, thanks alot man!

So for the Voltage equation, if one q is -2 microcouloumbs
then i minus the whole kq1/r term right

4. Jan 31, 2006

### SS2006

like i have 1, 18000 volts, the second was 9000 volts, and the third also 9000 volts but the coulumb charge was -2, instead of 2 like the first
so does that mean the final voltage is only 18000,
so a negative couloumb charge actually subtractsf rom the total voltage?

5. Feb 1, 2006

### DaMastaofFisix

Hmm, don't eaxctly see where you are going with this. You give voltage results and the chare, but no distance r....seems to me that ur missing something. Post the problem exactly as it's worded and i'll try and lead you into the right direction.