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enter image description here

I am not able to understand how the two uniform line charges will look like in space. It would be of great help if someone could draw the uniform line charges in cartesian coordinates and upload it here.

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This should probably get a homework tag, but here goes:

Since I can't give you a 3D image on a 2D screen, let's just take a slice. Luckily the problem is set up to do that reasonably easily. We can represent two at the same, z = +1 and z = -1. The equation for the sphere in A is $$(x-3)^2+(y-1)^2+z^2=2^2$$ for z = +/- 1 $$(x-3)^2+(y-1)^2+1=4$$ $$(x-3)^2+(y-1)^2=3$$ So we can draw a circle with center (3,1) and radius sqrt(3) in the z = +/-1 slice. Since y = 1 is the center, the y = 1, z=+/-1 lines will run right through the center of our slices and the total distance will be the diameter 2*sqrt(3) in both cases for a total distance of \$ 4 \cdot \sqrt3 \$ meters. Multiply by 20 for \$80 \cdot \sqrt3 \$ nC

Problem B is a little tougher enter image description here But with a bit of geometry we can show that the chord is 2sqrt(2) for each slice, thus a total charge of \$ 4 \cdot \sqrt2 \$ meters or \$ 80 \cdot \sqrt2 \$ nC

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  • \$\begingroup\$ oops, those are just the lengths in meters, so you need to multiply by 20 to get nC of charge \$\endgroup\$ – MikeP Jun 13 '20 at 12:00
  • \$\begingroup\$ 3rd formula should be "= 3". It's a good start but you need to apply the fixes and not leave it hanging. \$\endgroup\$ – Andy aka Jun 13 '20 at 12:17
  • \$\begingroup\$ good catch @Andyaka! I fixed it! \$\endgroup\$ – MikeP Jun 13 '20 at 15:15

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