- #1
MitsuShai
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1.
a) 0.102m
I set the electric fields equal to each other and canceled out the ks:
3q/(2d^2)= 8q/(d+x)^2 and then I solved from there.
b) 9.78*10^7
3q + 8q= 2.2e-5
E= 9e9 (2.2e-5/.045^2)= 9.78e7 N/C
c)1.56*10^-11
F=qE= 9.78e7*1.6e-19= 1.56e-11 N
2.A
3.D
4.D
5. I have no idea how to approach this problem.
Last Page: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/lastpg.jpg
6.D
7. I have no idea. Wouldn't it be parallel to both the +x and -x axes?
a) 0.102m
I set the electric fields equal to each other and canceled out the ks:
3q/(2d^2)= 8q/(d+x)^2 and then I solved from there.
b) 9.78*10^7
3q + 8q= 2.2e-5
E= 9e9 (2.2e-5/.045^2)= 9.78e7 N/C
c)1.56*10^-11
F=qE= 9.78e7*1.6e-19= 1.56e-11 N
2.A
3.D
4.D
5. I have no idea how to approach this problem.
Last Page: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/lastpg.jpg
6.D
7. I have no idea. Wouldn't it be parallel to both the +x and -x axes?
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