Recent content by macaco

thanks again TT

according to my most recent calculations, r = -1.79 x 10^-28m ?

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...

The one thing I did not understand, is why you would not use the negative symbols, and rank the negatives below the positives?

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...

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...

thanks so much guys! =] people might actually be able to read my equations now. LOL

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...

i hope those equations still make sense; it deleted the spaces in the bottom row. The first one should be; F= K Qq ----- d^2

No, we haven't touched on Gauss's law yet. I'm still not sure how to rank the equations after I've substituted the values. so far I have; q1=> F= K Qq ----- d^2 q2=> F= K 2Qq ------ 3d^2 q3=> F= K -3Qq...