Location A is 3.00 m to the right of a point charge q. Location B lies on the same line and is 4.70 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?

Question

Vb-Va=45= kq/4.7-kq/3

solve for q, you know k

Answer 1

45V =kq/rB - kq/rA
45V= k(q/4.70 - q/3.00)
45/k = q(1/4.70 - 1/3.00)

q= 45/(k(0.213-0.3330
= 45/((8.99E9)(-0.12)
= 45/(-1.08E-9)
= -4.17E-8C

Answer 2

Location A is 2.90 m to the right of a point charge q. Location B lies on the same line and is 3.50 m to the right of the charge. The potential difference VB - VA = 55.0 V. What is the magnitude and sign of the charge?

Vb-Va=55.0= kq/3.50-kq/2.90