Homework Help: Electric Potential and Second Charge

1. Sep 10, 2014

carpelumen

1. The problem statement, all variables and given/known data
The electric potential at a position located a distance of 18.6 mm from a positive point charge of 7.50×10-9C and 10.9 mm from a second point charge is 1.02 kV. Calculate the value of the second charge.

2. Relevant equations
q2 = r2 * (v - (k * q1 / r1)) / k

3. The attempt at a solution
v = 1.02 kV = 1.02 x 10^3 V = 1020 V
q1 = 7.50 x 10^-9 C
r1 = 18.6 mm = 0.0186 m
r2 = 10.9 mm = 0.0109 m
k = 9 x 10^9 Nm^2/C^2

q2 = (.0109) * (1020 - (9E9*7.5E-9/.0186)) / 9E9
q2 = -3.15E-9 nC

I keep getting the same answer but the homework system says it's wrong!

2. Sep 10, 2014

BvU

I get -3.16 10-9 C (not nC). Rounding off error ? Or is it the nC instead of the C ?

3. Sep 10, 2014

carpelumen

Wow. I feel like a total idiot. IT IS C instead of nC! I figured it couldn't be a unit error because the homework system usually notifies us of that.

Thanks so much!