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...
You're a legend Tiny Tim.
(The legend of Tiny Tim; sounds like a good book title =P)
Didn't think of taking out a common factor.
The charges in ascending order, according to the values left would be;
q2=> 0.66
q1=> 1
q4=> 1.33
q3=> 1.5
q5=> 2.5
(hopefully)
Thanks again TT
=]
Homework Statement
5 point charges;
q1; charge = +q distance = d
q2; charge = +2q distance = 3d
q3; charge = -3q distance = 2d
q4; charge = -4q distance = 3d
q5; charge = -5q distance = 2d
are placed in the vicinity of an insulating spherical shell with a charge (+Q), distributed...
I've been trying to edit one of my posts (to make the equations more eligible), but the 'edit' button doesn't appear?
Yet still does for my other threads?
Any idea why this would be?
Do threads time out after a few days?
And, yes, I am fairly bad with computers
=]
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...
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.
=]
This may be a stupid question, but is there a way to put spacing at the start of a line. Sim ilar to what the 'Tab' key would do in a word processor.
Just to make the equations I'm writing clearer.
=]
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?
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...
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...