Where should the proton be placed to balance F?

  • Thread starter macaco
  • Start date
  • #1
14
0

Homework Statement



A point charge q = -0.5nC is fixed at the origin. Where must a proton be placed in order for the electrical force acting on it to be exactly opposite its weight?


Homework Equations



F=
k(q1)(q2)
------------
r^2


The Attempt at a Solution


Trying to solve for distance (r), but coming across a few problems;

If the force acting on it is exactly opposite its weight, does this make F=0?

should I be substituting 0 for F?
Where would I substitute -0.5? in q1 or q2?
And what would I place in the other?

???
 

Answers and Replies

  • #2
76
0
A tips for the force: When you break out r you will need to take the square root, the value inside must be positive to yield a non-imaginary number. What does this say about the force?
It can't be zero, because then you would have to divide by zero, which gives a non-defined answer.

For the charges, it doesn't matter if the proton is q1 or q2, the same goes for the point charge. The expression won't change.
 
  • #3
2,479
100
KQ1Q2/r^2=mg
 
  • #4
14
0
Anden, I substitute 1 for q1 (the proton) and -0.35 x 10^-9 for q2 (the point charge)?

so I now have;
F=
K (1)(-0.35x10^-9)
---------------
r^2

Dadface, were you saying I should use F=m.g to find the value of force?
g would be 9.8,
what would m be?

would I use the mass of an electron (9.1×10^-31kg)
 
  • #5
14
0
the equation I now have is;
F= (9.1x10^-31)x(9.8). This equals 8.918x10^-30.
therefore;
8.918x10^-30=
(9x10^9)x(1)(-0.35x10^-9)
---------------
r^2

Then solve for r?
 
  • #6
76
0
A proton does not have the charge 1 C, that is the charge of 6,215 * 10^18 protons.

Dadface equation is almost right, with one modification "mg" needs to be replaced by "-mg", this is needed to ensure a positive result inside the square root.

The mass you must is the one for the proton, its mass is roughly 1840 times bigger than the electrons
 
  • #7
14
0
so F=(9.81)x(1.672x10^-27) => F=1.64x10^-26

the last thing I'm confused about is what to substitute for q1 and q1?
I know they are the point charges, and that one of them = -0.35x10^-9.
How do I figure out what the other point charge is?
Once I have that I should be good to go.

=]
 
  • #8
2,479
100
The charge on the proton is 1.6 times ten to the power of minus nineteen Coulombs
 
Last edited:
  • #9
76
0
Protons and electrons have the same charge, but with different signs. Protons are positive 1 e, and electrons are negative 1 e (-1 e). I gave you a hint for the charge of the proton, if 1 C equals 6,242E18 Protons (The value I posted above is wrong, sorry about that), what is the charge of one proton?
 
  • #10
14
0
The equation would now read;
(1.64x10^-26) = (8.99x10^9) (1.6x10^-19)(-0.35x10^-9)
------------​
r^2​
is that right?

rearrange the equation to make r the subject.

solve the equation, giving me;
r= -4.148x10^-46m
???

Hopefully I have it right. AT least I now have some sort of clue on how to solve it at the worst.
Thanks so much for the incredible patience Anden and Dadface.
I only see my physics teacher every thursday so if I get stuck on something it drives me insane.
 
  • #11
14
0
still a bit unsure on this;
q1 (the electron) has a charge of -3.5 x 10^-9 C
q2 (the proton) has a charge of 1.6 x 10^-19 C

to find F;
F = m.g
m of an electron is; 9.1 x 10^-31
g is; 9.8

therefore F= 8.9 x 10^-30 N

the equation will now be;
8.9 x 10^-30=
9 x 10^9 x (-3.5 x 10^-9)(1.6 x 10^-19)
----------------------​
r^2​
 
  • #12
14
0
according to my most recent calculations, r = -1.79 x 10^-28m

?????
 

Related Threads on Where should the proton be placed to balance F?

Replies
15
Views
615
Replies
3
Views
2K
Replies
2
Views
6K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
21
Views
1K
  • Last Post
Replies
3
Views
630
Top