Recent content by saxyliz
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Energy of a Dipole in a Constant E Field
Thank you so much! That really cleared it up for me!- saxyliz
- Post #3
- Forum: Introductory Physics Homework Help
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Energy of a Dipole in a Constant E Field
Homework Statement The graph shows the potential energy of an electric dipole which is in a constant electric field; only the electric force is acting on the dipole. Consider a dipole that oscillates between plus and minus 65 degrees. a) What is the dipole's mechanical energy? b) What...- saxyliz
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- Constant Dipole Energy Field
- Replies: 2
- Forum: Introductory Physics Homework Help
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Nonuniformly Charged Semicircle
Sorry! I'm totally new to this and didn't get what you meant in the first place (about the θ, etc.) Why isn't it λ_0 ? Wouldn't it be sin^2 then? I think I get what you mean about the components, give me a second. EDIT: I got it! Thanks so much for your help!- saxyliz
- Post #7
- Forum: Introductory Physics Homework Help
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Nonuniformly Charged Semicircle
So, dq = \lambda d \theta? Then I plug that into Coulomb's Law to solve for dE, but how do I solve for the components? To be more specific, what I get is: dE=k\lambda d \theta /R^{2} Which is the same as dE=k\lambda_{0} sin(\theta) d \theta /R^{2} Do I use a sector for the...- saxyliz
- Post #5
- Forum: Introductory Physics Homework Help
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Nonuniformly Charged Semicircle
Oh yes sorry! It's times! Silly me!- saxyliz
- Post #3
- Forum: Introductory Physics Homework Help
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Nonuniformly Charged Semicircle
Homework Statement A thin glass rod is a semicircle of radius R, see the figure. A charge is nonuniformly distributed along the rod with a linear charge density given by \lambda=\lambda_{0} + sin(\theta), where \lambda_{0} is a positive constant. Point P is at the center of the semicircle...- saxyliz
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- Charged
- Replies: 6
- Forum: Introductory Physics Homework Help