Net forces on a charge particle

AI Thread Summary
The discussion revolves around calculating net forces on charged particles, specifically focusing on the forces acting on a left positive charge due to two other charges. The calculations indicate that the left charge experiences an inward force, while the negative charge experiences no net force. The original poster expresses uncertainty about their conclusion, which was marked differently in a test, leading to a request for feedback from the community. Participants suggest verifying the correctness of the calculations and emphasize the importance of reading all options before making a choice. Ultimately, the poster is encouraged to appeal the test result without mentioning the forum.
pondzo
Messages
168
Reaction score
0

Homework Statement



I have attatched a snapshot explaining the question

Homework Equations



## \vec{F} = \frac{kq_{1}q_{2}}{r^2}\hat{r} ##

The Attempt at a Solution



First I will calculate the force on the left charge due to the other two

## \vec{F}_{net} = \frac{-kq^2}{d^2}(-1,0)+\frac{kq^2}{(2d)^2}(-1,0) = (\frac{3kq^2}{4d^2},0) ##
Which means the left positive charge will feel a force pulling it inwards, and through symmetry I can say that the right positive charge will be pulled inwards due to the force of the other charges.
Also for the negative charge; ## \vec{F}_{net} = \frac{-kq^2}{d^2}(-1,0)+\frac{-kq^2}{d^2}(1,0) = 0 ##
Based on this i would choose option D.

I was wondering if I have come to the right conclusion, if my calculations are correct, and if the way I approach the problem is correct. Thank you.
 

Attachments

  • Physics QN 1.PNG
    Physics QN 1.PNG
    22.8 KB · Views: 519
Physics news on Phys.org
You need reassurance ? PF isn't really meant for that.
Or do you want some tips for this kind of (low didactical quality) questions ?

First you have to check if only one answer is right, or that perhaps more than one (or, wow, none) is correct.
Here it clearly says ONE, so no problemo.

Second: try to be economical: if no calculations are needed, don't make them. Same charge further away = less F suffices. Or: twice as far, quarter the force.

Then: A and B are clearly nonsense. So is C (no calculating needed).
Elementary, Watson: D is left and it passes scrutiny.
Double check and move on to the next exercise.

[edit] and yes, your calculations are correct...
 
Well actually, this was a question on my physics test. And yes it was the easiest low quality question, which is why I posted it in the "introductory physics" section, just in case you missed that.

Anyway the point is, in the marking key, the solution was listed as E. So I wanted to see what the PF community thought before I made an appeal. But thanks for your comments anyway.
 
Ah, you saved a nice bombshell for later.

I'll add a tip: read all answers before embarking on choosing something. More or less as a reminder to myself (I missed E completeley :redface:, a reason to be modest instead of feeling insulted...).

Make your appeal ! And don't mention PF, it really isn't meant for that -- but please tell us how it went !
 
BvU said:
You need reassurance ? PF isn't really meant for that.
My understanding of the rules is that this kind of question is perfectly acceptable.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top