anonymous
  • anonymous
Two charges +4e and +2e are fixed at points (0,0)cm and (0,40)cm on the x axis. A third charge e is placed on the straight line joining the two charges so that e is in equilibrium. Find the position of e.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Algebraic! any clues
anonymous
  • anonymous
|dw:1348695082199:dw|
anonymous
  • anonymous
q1 (+4e) pushes a positive charge to the right. q2 (+2e) push a positive charge to the left. at some point between them those opposing pushes will be equal in magnitude. sub.s in x and .4-x for r1 and r2 and solve for x

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
is it clear?
anonymous
  • anonymous
not clear at all
anonymous
  • anonymous
the forces are equal in magnitude and opposite in direction.
anonymous
  • anonymous
| e*E1 |=| e*E2 | E1 = kq1/(r1)^2 E2 = kq2/(r2^2)
anonymous
  • anonymous
r1 is the distance from the charge at the origin to the location of the point where they 'test' charge is in equilibrium. That distance is x. r2 is the distance from the charge at the .4m to the location of the point where they 'test' charge is in equilibrium. That distance is .4m - x
anonymous
  • anonymous
E1 = kq1/(r1)^2 = E2 = kq2/(r2)^2 kq1/(r1)^2 = kq2/(r2)^2 kq1/(x)^2 = kq2/(.4-x)^2
anonymous
  • anonymous
solve the quadratic for x
anonymous
  • anonymous
clear now?
anonymous
  • anonymous
@Algebraic! not clear..how do you interepret Two charges +4e and +2e are fixed at points (0,0)cm and (0,40)cm..also how u derive the value for r1 and r2..what about the 3rd charge?
anonymous
  • anonymous
+4e charge is at x=0; +2e charge is at x=.4
anonymous
  • anonymous
r for the +4e charge is the distance from x=0 to the equilibrium point we're trying to find... call that distance x, r for the +2e charge is then the distance from x=.4 to the equilibrium point we're trying to find so .4-x
anonymous
  • anonymous
|dw:1348774582953:dw|
anonymous
  • anonymous
there is no 4 its 40..also the (0,0) isnt the x,y pairs?
anonymous
  • anonymous
.4
anonymous
  • anonymous
both charges are on the x axis and the equilibrium point is on a line between them, so it's on the x axis too. I don't know why they wrote it like this: "(0,0)cm and (0,40)cm on the x axis." it contradicts itself. so maybe check it to see if that's what the problem actually says. I assumed what they actually meant was both charges on the x axis one at x=0 one at x=.4
anonymous
  • anonymous
anyway, it's the same problem if those are actually meant to be ordered pairs, because: |dw:1348775056115:dw|
anonymous
  • anonymous
ok?
anonymous
  • anonymous
so u mean the 3rd charge is in the middle of both charge?
anonymous
  • anonymous
you have to locate where the third charge is... ...that's the point of the problem. but yes, it's probably on the x-axis, depending on what your problem statement actually says...
anonymous
  • anonymous
how do we solve this kind of quadratic 4e/x^2 = 2e/(4-x)^2
anonymous
  • anonymous
4e/x^2 = 2e/(.4-x)^2
anonymous
  • anonymous
cross multiply
anonymous
  • anonymous
4e(.4-x)^2 = 2ex^2
anonymous
  • anonymous
x is 2.3
anonymous
  • anonymous
@Algebraic! do you believe i just saw the .4 and not 4...thanks a lot

Looking for something else?

Not the answer you are looking for? Search for more explanations.