How Do You Calculate the Charge Between Two Points Given Voltage Difference?

[SilentBlade91]

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.

 

[collinsmark]

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)

Try again!

1/r_B - 1/r_A ≠ 1/(r_B - r_A)

You need to find a common denominator.