- #1

MitsuShai

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1.

a) 0.102m

I set the electric fields equal to each other and cancelled 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 [Broken]

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 cancelled 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 [Broken]

6.D

7. I have no idea. Wouldn't it be parallel to both the +x and -x axes?

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