Physics: Coulombs Law: Two small, positively charged spheres have a combined charge of 13.0 × 10-5 C+++?

Question

If each sphere is repelled from the other by an electrostatic force of 0.600 N when the spheres are 1.80 m apart, what is the charge on the sphere with the smaller charge?

Answer 1

Favourite answer

0.600 N = (9 x 10^9 Nm^2/C^2)*Q*(1.3 x 10^(-5)C - Q) / (1.80 m)^2.

Agreeing that "Q" will come out in coulombs, let's abandon the units for a moment.

0.600 = (9 x 10^9) Q*(1.3 x 10^(-5)C - Q) / 1.80^2 =>

1.944 = (9 x 10^9) Q*(1.3 x 10^(-5) C - Q) =>

(9 x 10^9)Q^2 - 1.17 x 10^5 Q + 1.944 = 0.

Now you have a quadratic equation to be solved for Q.

Use the quadratic formula, that's what it's for !

Answer 2

13.0 × 10-5 = 130–5 = 125 C

strange number,,,

what does +++ indicate?

F = kQ₁Q₂/r² = 0.8

Q₁Q₂ = 0.8•1.8²/9e9 = 2.88e-10

Q₁+Q₂ = 125

Q₁ = 125 – Q₂

Q₁Q₂ = 2.88e-10

(125 – Q₂)Q₂ = 2.88e-10

Q₂²  – 125Q₂ + 2.88e-10 = 0

quadratic equation:to solve ax² + bx + c = 0x = [–b ± √(b²–4ac)] / 2aQ₂= [125 ± √(125²–4•2.88e-10)] / 2

Q₂= [125 ± 124.999999999983] / 2

Q₂ = 125 C

Q₁ = 8.6e-12 C

Coulomb's law, force of attraction/repulsion

F = kQ₁Q₂/r²

   Q₁ and Q₂ are the charges in coulombs

   F is force in newtons

   r is separation in meters

   k = 8.99e9 Nm²/C²