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SilentBlade91
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Homework Statement
Location A is 3.04m to the right of a point charge q.
Location B lies on the same line and is 3.87m to the right of
the charge. The potential difference between the two locations
is VB - VA = 45.0V . What is the magnitude and sign of the
charge?
Homework Equations
V=(kq)/r I think
The Attempt at a Solution
Using the equation above I modified it to
(VB-VA)=(kq)/(rB-rA)
and solved for q=((VB-VA)(rB-rA))/k
so q=((45.0V)(3.87m-3.04m))/9.0*10^9 N*m^2/C^2
and got q=4.15*10^-9 C but it wasnt right. Any help is greatly appreciated.