Answer to Question #89057 in Electricity and Magnetism for Promise Omiponle

Question

1. Three point charges are arranged along the x-axis. Charge q1 = +5.00µC is at the origin, and charge q2 = -8.00µC is at x = 0.400m. Charge q3 = -10.00µC. Where is q3 located if the net force on q1 is 9.00N in the -x-direction(negative x direction)?

2. An 18-gauge copper wire, with a diameter of 2mm carries a constant current of 1.86A to a 240-W lamp. The free-electron density in the wire is 6.5x10^28 per cubic meter and the resistivity is 1.72x10^-8 Ωm. Find:
a. the curremt density
b. the drift speed
c. the electric-field magnitude in the wire
d. the potential difference between two points in the wire 60m apart
e. the resistance of a of this wire.

Answer 1

1.

$$F_{12}=\frac{kq_1q_2}{r_{12}^2}=\frac{(8.99 \cdot 10^{9})(5 \cdot 10^{-6})(8 \cdot 10^{-6})}{0.4^2}=2.2475 \N$$

$$F_{13}=F-F_{12}=-9-2.2475=11.2475 \ N$$

$$F_{13}=-\frac{kq_1q_3}{r_{13}^2}=-\frac{(8.99 \cdot 10^{9})(5 \cdot 10^{-6})(10 \cdot 10^{-6})}{r_{13}^2}=-11.2475 \N$$

$$r_{12}=0.200\ m$$

in the -x-direction (negative x direction).

$$x=-0.200 \ m$$

2. a)

$$j=\frac{I}{\pi r^2}=\frac{1.86}{\pi (0.001)^2}=5.92 \cdot 10^5\frac{A}{m^2}$$

b)

$$v=\frac{j}{ne}=\frac{5.92 \cdot 10^5}{(6.5 \cdot 10^{28})(1.6 \cdot 10^{-19})}=5.7 \cdot 10^{-5}\frac{m}{s}$$

c)

$$E=j\rho =(5.92 \cdot 10^5)(1.72 \cdot 10^{-8})=0.0102 \frac{V}{m}$$

d)

$$V=Ed=(0.0102)(60)=0.612\ V$$

e)

$$R=\rho \frac{d}{A}=(1.72 \cdot 10^{-8}) \frac{60}{\pi (0.001)^2}=0.328 \ \Omega$$