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Electric field of Plastic Pipes

A plastic pipe has an inner radius of a = 31.00 cm and an outer radius of b = 68.00 cm. Electric charge is uniformly distributed over the region a < r < b. The charge density in this region is 42.00 C/m3. Calculate the magnitude of the electric field at r = 0.29 m.
Calculate the magnitude of the electric field at r = 0.57 m. Calculate the magnitude of the electric field at r = 1.73 m.

I thought if I did E= lambda/2*pi*Eo*r, and placed the r asked I can solve this. But that was wrong. Why is this wrong?

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A plastic pipe has an inner radius of a = 31.00 cm and an outer radius of b = 68.00 cm. Electric charge is uniformly distributed over the region a < r < b. The charge density in this region is 42.00 C/m3. Calculate the magnitude of the electric field at r = 0.29 m.
Calculate the magnitude of the electric field at r = 0.57 m. Calculate the magnitude of the electric field at r = 1.73 m.

I thought if I did E= lambda/2*pi*Eo*r, and placed the r asked I can solve this. But that was wrong. Why is this wrong?

Answer:
Following figures show vertical and horizontal cross sections of the pipe. The shaded area shows the region in which electric charge is uniformly spread.
The charge is spread over the entire volume of the pipe and hence, we must calculate the total charge per unit length by considering the volume of the unit length of the pipe material as follows.

Electric ...

Solution Summary

The magnitude of the electric field is calculated at different radius'. The magnitude of the electric field is determined.

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