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 2 years 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.
 2 years 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.

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woleraymond
 2 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! any clues

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

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1q1 (+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

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

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

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1r1 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

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

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

woleraymond
 2 years ago
Best 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?

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

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1r 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

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

woleraymond
 2 years ago
Best ResponseYou've already chosen the best response.0there is no 4 its 40..also the (0,0) isnt the x,y pairs?

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1both 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

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

woleraymond
 2 years ago
Best ResponseYou've already chosen the best response.0so u mean the 3rd charge is in the middle of both charge?

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1you 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...

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

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.14e/x^2 = 2e/(.4x)^2

woleraymond
 2 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! do you believe i just saw the .4 and not 4...thanks a lot
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