1) An insulating rod of length L=8.15 cm is uniformly charged with q=-4.23 fC. (a) What is the linear charge density of the rod? (b) what is the electric field at point P (it's a vector, so both magnitude and direction!). P is a=12.0 cm away from the end of the rod. Work it out! Don't just quote the special case formula, I want to see you you do it from scratch. (c) what is the answer if "a" was 50 m instead? (d) at 50 m, what if it wasn't a rod but a poínt charge with the same "q"? -4 P L 2) You've got an electric dipole in an electric field of magnitude 46.0 N/C. You want to flip it 180 degrees from where it started (it starts pointing at an angle of 64 degrees from the electric field). The electric dipole moment of this thing is 3.02x10-25 Cm (magnitude: direction described above). How much work do you do while cranking it around?

Principles of Physics: A Calculus-Based Text
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ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
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Chapter19: Electric Forces And Electric Fields
Section: Chapter Questions
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please do question #2 and if you write the explanation next to the steps while you're doing it that would be great.

1) An insulating rod of length L=8.15 cm is uniformly charged with q=-4.23 fC. (a) What is the linear charge
density of the rod? (b) what is the electric field at point P (it's a vector, so both magnitude and direction!). P is
a=12.0 cm away from the end of the rod. Work it out! Don't just quote the special case formula, I want to see
you you do it from scratch. (c) what is the answer if "a" was 50 m instead? (d) at 50 m, what if it wasn't a rod but
a point charge with the same “q"?
-9
P
L
a
2) You've got an electric dipole in an electric field of magnitude 46.0 N/C. You want to flip it 180 degrees from
where it started (it starts pointing at an angle of 64 degrees from the electric field). The electric dipole moment of
this thing is 3.02x10-25 Cm (magnitude: direction described above). How much work do you do while cranking it
around?
Transcribed Image Text:1) An insulating rod of length L=8.15 cm is uniformly charged with q=-4.23 fC. (a) What is the linear charge density of the rod? (b) what is the electric field at point P (it's a vector, so both magnitude and direction!). P is a=12.0 cm away from the end of the rod. Work it out! Don't just quote the special case formula, I want to see you you do it from scratch. (c) what is the answer if "a" was 50 m instead? (d) at 50 m, what if it wasn't a rod but a point charge with the same “q"? -9 P L a 2) You've got an electric dipole in an electric field of magnitude 46.0 N/C. You want to flip it 180 degrees from where it started (it starts pointing at an angle of 64 degrees from the electric field). The electric dipole moment of this thing is 3.02x10-25 Cm (magnitude: direction described above). How much work do you do while cranking it around?
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