Posted by **CombatVetUSMC** on Saturday, July 14, 2012 at 1:03am.

A positive charge +q1 is located to the left of a negative charge -q2. On a line passing though the two charges, there are two places where the total potential is zero. The first place is between the charges and is 4.03 cm to the left of the negative charge. The second place is 7.01 cm to the right of the negative charge. (a) What is the distance between the charges? (b) Find q1/q2, the ratio of the magnitudes of the charges.

- Physics -
**Elena**, Saturday, July 14, 2012 at 7:01am
x1=4.03 cm, x2=7.01 cm.

d=?

The left point is A, the right point is B.

Potential due to the point charge is

φ = k•q/r.

Then, at the point A

φ(A1) =k•q1/(d-x1),

φ(A2) = -k•q2/x1,

φ(A1)+ φ(A2)=0 =>

k•q1/(d-x1) =k•q2/x1,

q1/q2=(d-x1)/x1….. (1)

At the point B

φ(B1) =k•q1/(d+x2),

φ(B2) = -k•q2/x2,

φ(B1)+ φ(B2)=0 =>

k•q1/(d+x2) =k•q2/x2,

q1/q2=(d+x2)/x2 …..(2)

Equate the right sides of the equations (1) and(2)

(d-x1)/x1 = (d+x2)/x2

d=2•x1•x2/(x2-x1) =

=24.03•7.01/(7.01-4.03) =18.96 cm.

q1/q2 =(d-x1)/x1 =

=(18.96-4.03)/4.03 = 3.7

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