Recent content by vampyric
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Volume charge density across a potential difference
Yes! Perfect sense! Cannot thank you enough for sharing your wisdom!- vampyric
- Post #9
- Forum: Advanced Physics Homework Help
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Volume charge density across a potential difference
Wow that is a beautiful solution, cancelling out at its finest! Just for further understanding, why am I able to use the charge for a proton to find the velocity but then when I am finding the volume charge density I have to use Q=LI/v? Is it right to take this second Q to be the total...- vampyric
- Post #7
- Forum: Advanced Physics Homework Help
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Volume charge density across a potential difference
Whoa good spotting! I used the given diameter :/ ahh so wrong. I honestly have no idea what to use here as I'm so confused about its shape. Whether I'm looking at the shape of the beam or the shape of the p.d. That makes sense considering it is a beam. I have length but I do not have a...- vampyric
- Post #5
- Forum: Advanced Physics Homework Help
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Volume charge density across a potential difference
Hi thanks for replying! So given the values for mass and charge of a proton (m=1.67×〖10〗^(-26) kg,Q=1.6×〖10〗^(-19) C), velocity can be determined: QV=(1.6×〖10〗^(-19) )(10×〖10〗^3 ) =1.6×〖10〗^(-15) J Then using QV=E=KE I was able to find a velocity, E=1/2 mv^2 v=√(2E/m) =1.38×〖10〗^6...- vampyric
- Post #3
- Forum: Advanced Physics Homework Help
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Volume charge density across a potential difference
Homework Statement A 1.0 μA proton beam is accelerated across a potential difference of 1.0 kV. Assume the beam has uniform current density over a diameter of 2.0 mm, and zero outside. Find: volume charge density in the beam, (HINT use λ=I/v where λ= charge/ unit length) The radial...- vampyric
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- Charge Charge density Density Difference Potential Potential difference Volume
- Replies: 9
- Forum: Advanced Physics Homework Help
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Understanding Vector Spaces with Mysterious Coefficients
Ohhhhh right. This is actually making sense now. Thank you so much! As for Nevermind, I was interpreting it to apply to all ordinary + as well and was confused as to why I hadn't seen this in R^3. I will never underestimate the significance of notation again :) Thanks!- vampyric
- Post #8
- Forum: Calculus and Beyond Homework Help
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Understanding Vector Spaces with Mysterious Coefficients
So taking this with respect to axiom 1: v+w=w+v for all v,w ∈ V (x',y')+(x,y)=(x'+x+1,y'+y+1) Is this correct?- vampyric
- Post #5
- Forum: Calculus and Beyond Homework Help
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Understanding Vector Spaces with Mysterious Coefficients
Sorry about that, I was in a rush to get to class. So I'm not sure if I get this For an addition operation (x,y)+(x',y') from definition you just simply add +1 to each coefficient...is this what you mean? I haven't noticed it being done for the case (x,y,z), is this because it is in R^3...- vampyric
- Post #4
- Forum: Calculus and Beyond Homework Help
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Understanding Vector Spaces with Mysterious Coefficients
Homework Statement Hi there, I'm very new to vector spaces and just can't seem to figure this one problem out. The question ask's to determine if (V,+,*) is a vector space. I am given V=R^2 (x,y)+(x',y')=(x+x'+1,y+y'+1) for addition on V and λ*(x,y)=(λx+λ-1,λy+y-1) (λ∈ℝ) for...- vampyric
- Thread
- Vector Vector spaces
- Replies: 12
- Forum: Calculus and Beyond Homework Help