# columbs law 3 charges

Hi,

I know coloumbs law but when 3 charges are involved i am a little unsure.
F=kq1q2/r^2

if I have 3 charges in a row
R  (+q charge)
S  (+2q charge)
T  (+3q charge)

distance between R,S =d1
distance between S,T =d2 which looks 3 times distance as d1

q1)to work out 0 new electric force when d2 is = to...arent I just working out if d1 or d2=0? The answer is sqroot 3 *d1
q2) using a -2q charge and touching  R,S,T  what is the charge of T?
Isnt this just stay as +3Q? the answer is 15q/8
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Author Commented:
wait I worked out q1 as the force is = on both sides so d2=sqroot 3 *d1
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Commented:
q1)to work out 0 new electric force
Do you mean 0 net force on S?

q2) using a -2q charge
using it for what?
Do you mean take an object U starting with charge -2q and bring it into contact with R then into contact with S, then into contact with T?
Are all the objects equally sized conductive spheres?
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Commented:
You need to be a little more specific as to just what you are asking. The placement of the charges are all right. (But I cannot see why you say ("d2 which looks 3 times distance as d1").
I assume that the original question asks for the electric field at some point on the line formed by the charges. Am I correct? If so, where?  Are you looking for the location(s) where the field = 0? What values, if any are atatched by the problem to d1 and d2?
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Author Commented:
The length of d2 has been solved as it is the square root of 3 X d1
no values are attached to d1,d2.

What I need is q2) using a -2q charge and touching  R,S,T  what is the charge of T?
Isnt this just stay as +3Q? the answer is
15Q/8
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Author Commented:
yes to ozo
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Commented:
If R,S and T are equal size conductive spheres, and we introduce a new conductive sphere of the same size and initial charge -2q, then if we bring it into contact with R, the charges would equalize at -q/2 each.  If  the new object is then brought  into contact with S, charges would equalize at +3q/4  each.  If the new object is then brought into contact with T, the charges would equalize at 15q/8.
Is that what you mean?
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Author Commented:
"if we bring it into contact with R, the charges would equalize at -q/2 each"
how?
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Commented:
Symmetry.

If the charge on identical spheres were different, there would be  a net field.
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Author Commented:
I dont get how you get -q/2?
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Commented:
+q  + -2q = -q/2 + -q/2
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