Master Point Charges with Ease: Homework Equations & Attempt at Solution

In summary: They are having trouble understanding where to start and are lost after trying a few different things. After some algebra, they are able to find the co-ordinates for the unknown charge and solve for it.
  • #1
pat666
709
0

Homework Statement


See attachment--please!


Homework Equations



F=kq1q2/r^2

The Attempt at a Solution


I can't see where to start this question, I can do questions where there is three point charges and it asks you to find the net force on one of them and i think that in essence this question is the same. the problem is that nothing i do/think of seems to work. I thought that F1x + F2x =0 might work but i always end up with two variables (r and qx). Please help
 

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  • #2
1. Forces are vectors. Forces in the same direction can be added together.
All three charges lie on the x-axis, including your unknown charge (why?, hint: draw diagrams showing electric field vectors of positive charges)

2. what is the only region where you can put this charge so the forces can potentially add up to zero?
 
  • #3
at some point in the middle I am thinking but i was trying to find the co-ordinates - is that not what part a is after? if so I still can't see how to do part b. thanks ush
 
  • #4
pat666 said:
at some point in the middle
yes.
You know the total distance between the two charges is 0.9m
let "a" be the distance x=0 to the unknown charge
now, without adding another variable, what is the distance from the unknown charge to x=0.9 ?
 
  • #5
how can i work that out??- sorry but i am now thoroughly confused...
 
  • #6
lol.
if 'a' is the distance from x=0 to the unknown charge, then '0.9 - a' must be the distance from the unknown charge to x=0.9

why?

if you add "a" + "0.9 - a" you get your total distance "0.9". so 'a' is somewhere within your total distance.

knowing this.. you can plug in these variables for your radius. ("a" being the radius from 5.0 μC to the unknown charge, and "0.9-a" being the radius from the 8.0 μC to the unknown charge).

solving for "a" is now strictly algebraic
--
Recall: the two forces exerted on the point cancel out, therefore the force exerted on the unknown point by the 5.0 μC charge cancels with the 8.0 μC charge on your free body diagram (which I highly recommend drawing at the very beginning).
 
  • #7
thankyou! Ill post back if i have any more issues- thanks!
 
  • #8
hey ush, I've been trying to solve this for ages now and i keep getting -1.5m and that can't be right. I am not sure what i am doing wrong??
 
  • #9
if i minus them instead of +ing them i get 0.397m? - i really need help..
 
  • #10
Put down your solution here. I'll check it.
 
  • #11
thanks graphene
F1x+F2x=0
0=9E9*5E-6/a^2+-9E9*8E-6/(0.9-a)^2
then I solve for a using the 89 and get 0.397 or -1.5 depending on plus or minus in the middle
 
  • #12
So, a = 0.397 is the only possible solution.

That gives you the the position.

Now you'll have to find out what charge you need to place there so that the net force on the first two charges is zero.
What should be the sign of the third charge so that the net force is zero on the first two charges after the third charge is placed?
 
  • #13
-1.5 is a distance that's NOT in-between the two charges, 0.397 is between the two charges. You concluded earlier that the unknown charge must be in-between the two charges.
 
  • #14
Hey ush sorry for the late reply- I have solved the problem - just wondering if you could check it please... I got d=0.397m and Q_3=-1.54*10^(-6) C
 
  • #15
pat666 said:
Hey ush sorry for the late reply- I have solved the problem - just wondering if you could check it please... I got d=0.397m and Q_3=-1.54*10^(-6) C

Your distance is correct.
You're charge for Q3 is incorrect. Q3 can be any charge it wants to be.. it does not matter because on the free body diagram of Q3 the only forces acting on it are Q1 and Q2. Q3 cannot exert a force on itself therefore it makes no difference what Q3 is.

Unless I'm missing something from the question- I don't believe Q3 has to be any particular magnitude of charge.
 
  • #16
hey Ush - if you look at question b it does want both the magnitude and sign
 
  • #17
Q3 cannot be any charge.
The questions requires that the forces on q1 & q2 be zero as well. So, you'll need to place a particular charge for q3.

pat666 - again, if you could show me your solution, i'll verify it.
 
  • #18
I got d=0.397m and Q_3=-1.54*10^(-6) C
thanks again graphene very very helpful
 
  • #19
graphene said:
Q3 cannot be any charge.
The questions requires that the forces on q1 & q2 be zero as well. So, you'll need to place a particular charge for q3.

pat666 - again, if you could show me your solution, i'll verify it.

You're right, I misread the question. I thought it was referring to Q3 only, but it says all charges.

In that case, a charge of -1.55E-6 should balance the forces, so yeah, that's right
 
  • #20
Thanks for all your help
 

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