Answer to Question #208412 in Electric Circuits for Kuina

Question

2. A point charge q₁ = 4.00 nC is placed at the origin, and a second point charge q₂ = -3.00 nC is placed on the x-axis at x= +20.0 cm. A third point charge q₃ = 2.00 nC is to be placed on the x-axis between q₁ and q₂. Let the potential energy of the three charges be zero when they are infinitely far apart.

a) what is the potential energy of the system of the three charges if q₃ is placed at x= +10.0 cm?

b) where should q₃ be placed to make the potential energy of the system equal to zero?

Answer 1

(a) Potential energy (U) is given by

$U=k(\frac{q_1q_2}{x_{12}}+\frac{q_1q_3}{x_{13}}+\frac{q_2q_3}{x_{23}}) \\ k=8.9 \times 10^9Nm^2/C^2\\ q_1=4nC=4\times 10^{-9}C\\ q_2=-3nC=-3\times 10^{-9}C\\ q_3=2nC=2\times 10^{-9}C\\ x_{12}=20cm=0.2m\\ x_{13}=10cm=0.1m\\ x_{23}=10cm=0.1m\\ U=8.9\times 10^{9}(\frac{4\times 10^{-9}\times -3 \times 10^{-9}}{0.2} \\+\frac{4\times 10^{-9}\times 2\times 10^{-9}}{0.1} \\+\frac{-3\times 10^{-9}\times 2\times 10^{-9}}{0.1}) \\ U=8.9\times 10^9(-6\times10^{-17}\\+8\times 10^{-17}-6\times10^{-17})\\ U=8.9\times10^9 \times10^{-17}(-4)\\ U=-35.6\times 10^{-8}\\ U=-356nJ\\$

(b)

$U=k(\frac{q_1q_2}{x_{12}}+\frac{q_1q_3}{x_{13}}+\frac{q_2q_3}{x_{23}})\\ 0=10^{-18}(\frac{4\times -3}{0.2}+\frac{4\times 2}{x} +\frac{-3\times 2}{0.2-x})\\ 0=-60+\frac{8}{x}-\frac{6}{0.2-x}\\ 60=\frac{8}{x}-\frac{6}{0.2-x}\\ 60=\frac{8(0.2-x)-6(x)}{x(0.2-x)}\\ 60=\frac{1.6-8x-6x}{0.2x-x^2}\\ 60=\frac{1.6-14x}{0.2x-x^2}\\ 60(0.2x-x^2)=1.6-14x\\ 12x-60x^2=1.6-14x\\ 60x^2-30x+1.6=0$

Solving the quadratic equation

x=0.06, 0.44

The distance is 0.06m(6cm) since it is between 0 and 20cm

System