Question

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.

Answer 1

@Algebraic! any clues

Answer 2

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

Answer 3

is it clear?

Answer 4

not clear at all

Answer 5

the forces are equal in magnitude and opposite in direction.

Answer 6

| e*E1 |=| e*E2 |
E1 = kq1/(r1)^2
E2 = kq2/(r2^2)

Answer 7

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

Answer 8

E1 = kq1/(r1)^2 = E2 = kq2/(r2)^2
kq1/(r1)^2 = kq2/(r2)^2
kq1/(x)^2 = kq2/(.4-x)^2

Answer 9

solve the quadratic for x

Answer 10

clear now?

Answer 11

@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?

Answer 12

+4e charge is at x=0; +2e charge is at x=.4

Answer 13

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