Location A is 3.40 m to the right of a point charge q. Location B lies on the same line and is 4.50 m to the right of the charge. The potential difference between the two locations is VB - VA = 45.0 V. What is the magnitude and sign of the charge?

Answer 1

ΔV=VB-VA =kq/r(B)-kq/r(A)=
=kq{r(a)-r(B)}/r(A)•r(B).

q = r(A)•r(B)•ΔV/k{r(a)-r(B)}=
=(3.4-4.5)•45/9•10⁹•3.4•4.5 =
= - 9.5•10⁻¹⁰ C