Need Help With Physics 2 Questions

Question

Three point charges are arranged along the x-axis. Charge q1= -5.00nc is located at x= 0.300m, and charge q2= -3.50nC is at x= -0.400m. Where along the x-axis can a positive charge q3 be placed and the net force on it be zero, other than the trivial answers of x= +infinity and x= -infinity.

Answer 1

Draw the arrangement of charges.

*------------------o------------* o is origin

q2 q1

|......x.....|..........0.7-x.......|

q1 = -5.00 nC

q2 = -3.50 nC

Per Static Electricity Forces, +/- or -/+ attracts and +/+ or -/- repels.

The positive charge placed in between the two negative charges will be attracted

in opposite direction.

Per Coulomb Law, vec (F) = - k (q'q"/r²) vec (F) indicates a vector quantity.

the distance between q1 and q2 is 0.300m - (-0.400m) = 0.700 m.

since |q1| > |q2|, so the equilibrium point should be closer to q2 than q1.

Let's set it to x meter from q2 and (0.7-x) meter to q1. see above drawing

Now setup the equation using Coulomb's Law:

F_q2q+ = k(q2q+ / x²) = k[q1q+ / (0.7 - x)²] = F_q1q+ q+ is the positive charge used

cancel k and q+ on both sides, we get

q2/x² = q1/(0.7 - x)²

put in values for q1 abd q2 (- sign and nC unit cancel each other too!)

3.5/x² = 5.0/(0.7 - x)²

cross multiply and simplify to get:

1.5 x² + 4.9x - 1.715 = 0

solve for x and get: 3.59 m and 0.319 m, since x should be less than 0.700 m

take 0.319 m to calculate coordinate for the equilibrium point.

To find the coordinate, x_i

x_i = - 0.400 m + 0.319 m

= - 0.081 m answer

Hope it helps.