Asked by Emma
Two fixed charges, -4μC and -5μC, are separated by a certain distance. If the charges are separated by 20cm, what is the magnitude of the electric field halfway between the charges?
My textbook says that the formula for electric field is (kq)/r^2.
But Since I have two charges (and both are negative) I don't really know what to put for q.
Would I solve for the electric fields separately and then find their vector sum to get the halfway point electric field?
Or would q be the difference between the two q's?
Please help lead me in the right direction!!! Thank you!
My textbook says that the formula for electric field is (kq)/r^2.
But Since I have two charges (and both are negative) I don't really know what to put for q.
Would I solve for the electric fields separately and then find their vector sum to get the halfway point electric field?
Or would q be the difference between the two q's?
Please help lead me in the right direction!!! Thank you!
Answers
Answered by
Damon
Well, you can find the electric field E due to charge 1 at charge 2, then multiply that field by charge 2 to find the repulsive force (note, same sign charge repels, different signs attract)
HOWEVER, normally you use Coulombs Law
F = k Q1 Q2 / r^2
( Google "Coulomb's Law" )
as you can see, it amounts to the same thing because the Electric field E is in fact the force on a charge of 1 coulomb.
HOWEVER, normally you use Coulombs Law
F = k Q1 Q2 / r^2
( Google "Coulomb's Law" )
as you can see, it amounts to the same thing because the Electric field E is in fact the force on a charge of 1 coulomb.
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