Question #89057
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.
1
Expert's answer
2019-05-06T10:11:05-0400

1.


F12=kq1q2r122=(8.99109)(5106)(8106)0.42=2.2475NF_{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

F13=FF12=92.2475=11.2475 NF_{13}=F-F_{12}=-9-2.2475=11.2475 \ N

F13=kq1q3r132=(8.99109)(5106)(10106)r132=11.2475NF_{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

r12=0.200 mr_{12}=0.200\ m

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


x=0.200 mx=-0.200 \ m

2. a)


j=Iπr2=1.86π(0.001)2=5.92105Am2j=\frac{I}{\pi r^2}=\frac{1.86}{\pi (0.001)^2}=5.92 \cdot 10^5\frac{A}{m^2}

b)


v=jne=5.92105(6.51028)(1.61019)=5.7105msv=\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=jρ=(5.92105)(1.72108)=0.0102VmE=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 VV=Ed=(0.0102)(60)=0.612\ V

e)


R=ρdA=(1.72108)60π(0.001)2=0.328 ΩR=\rho \frac{d}{A}=(1.72 \cdot 10^{-8}) \frac{60}{\pi (0.001)^2}=0.328 \ \Omega


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS