# Coulomb's Law and conversion of nano-columbs/coulombs

This is not a homework problem. I am working ahead for my Electricity and Magnetism course for next quarter and this is a Chapter 25 video tutor solution question pearson put out where they do a short video alongside a problem.

A rod with charge + 350 nC is being used to levitate a charged balloon, which has mass 3.0 g. The balloon is being held stationary 15 cm below the charged rod. What is the approximate charge on the balloon?

What I know: Fnet=0 because it's stationary and the balloon is negatively charged because the rod is positively charged. I know q1, r and mass of the balloon.

$$Fnet = F_g -F_b$$ $$mg = F_b$$ By using coulombs law I get an expression: $$mg = \frac{K_e\lvert q_1q_2\rvert}{r^2}$$ solving for q2 $$q_2 = \frac{mgr^2}{K_eq_1}$$ Now Ke is in coulombs so during this step I convert q1 to coulombs $$q1 = 350 *10^{-9}$$

This is in coulombs and I need to convert back to nano coulombs so I multiply this answer I've found by:

$$\frac{mgr^2}{K_eq_1}*10^{9} = 210 nC$$

After this point I need to assess my model and find the direction of q2. I said before the balloon was negatively charged so it's q should be negatively charged. Giving me an answer of -210 nC.

This is very close to the answer Pearson got but according to the video I am off by a factor of 10. They had 21 or 20 nC(they rounded to 20 without giving explanation why).

I am very confused. I have done all my work multiple times and even checked it on wolfram alpha.

I really want to build a good understanding of this chapter as these are the fundamentals of E&M and this course terrifies me a bit

Might you assist me with this somehow?

EDIT: After further exploration of the problem, I am almost certain Pearson forgot to multiply by g. Thank you Costrom for the feedback. I will be contacting Pearson, linking to this thread. Thanks for the feedback.

• You conversion from nC to C is off, but happens to cancel with a similar error in the next step. You should always have a smaller number for C than nC (it should be $10^{-9}$ not $10^9$) – tmwilson26 Dec 11 '15 at 17:09
• Yeah I fixed that. That was just an error of entering into stack exchange properly. It's updated. – jake mckenzie Dec 12 '15 at 17:32

In "normal" physics and engineering problems, I always try to use the base units to be extra careful (kg,m,C...)

so

$q_1 = 350\cdot10^{-9}C$, $r = 0.15m$ , $m=0.003kg$, $g = 9.81 \frac{m}{s^2}$

$q_2 = -\frac{mgr^2}{K_eq_1} = -\frac{0.003 \cdot 9.81 \cdot (0.15)^2}{8.987\cdot10^{9}\cdot350\cdot10^{-9}} \approx -210 nC$