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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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woleraymondBest ResponseYou've already chosen the best response.0
@Algebraic! any clues
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
dw:1348695082199:dw
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
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 .4x for r1 and r2 and solve for x
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
the forces are equal in magnitude and opposite in direction.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
 e*E1 = e*E2  E1 = kq1/(r1)^2 E2 = kq2/(r2^2)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
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
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
E1 = kq1/(r1)^2 = E2 = kq2/(r2)^2 kq1/(r1)^2 = kq2/(r2)^2 kq1/(x)^2 = kq2/(.4x)^2
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
solve the quadratic for x
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
@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?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
+4e charge is at x=0; +2e charge is at x=.4
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
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 .4x
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
dw:1348774582953:dw
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
there is no 4 its 40..also the (0,0) isnt the x,y pairs?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
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
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
anyway, it's the same problem if those are actually meant to be ordered pairs, because: dw:1348775056115:dw
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
so u mean the 3rd charge is in the middle of both charge?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
you have to locate where the third charge is... ...that's the point of the problem. but yes, it's probably on the xaxis, depending on what your problem statement actually says...
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
how do we solve this kind of quadratic 4e/x^2 = 2e/(4x)^2
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
4e/x^2 = 2e/(.4x)^2
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
@Algebraic! do you believe i just saw the .4 and not 4...thanks a lot
 one year ago
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