1
$\begingroup$

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?

Here is a link to the final answer that Pearson got:

Chapter 25 Video Tutor Solution

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.

$\endgroup$
  • $\begingroup$ 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 $10^9$) $\endgroup$ – tmwilson26 Dec 11 '15 at 17:09
  • $\begingroup$ Yeah I fixed that. That was just an error of entering into stack exchange properly. It's updated. $\endgroup$ – jake mckenzie Dec 12 '15 at 17:32
1
$\begingroup$

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

so

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

using your equation:

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

It appears that the Pearson answer is off... Is there any step in the video you mentioned that does not get the same intermediate answer as you?

$\endgroup$
  • $\begingroup$ I updated the original post with their answer and what they wrote. They don't show their work to get to -21 nC $\endgroup$ – jake mckenzie Dec 11 '15 at 18:17
  • $\begingroup$ looks like someone lost a decimal place during some mental math, then $\endgroup$ – costrom Dec 11 '15 at 19:08
  • $\begingroup$ Thanks costrom. Pearson is generally really good. I trusted my answer and my work but I was quite surprised they made a mistake here. Considering its a produced video. $\endgroup$ – jake mckenzie Dec 11 '15 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.