Recent content by Molderish
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Electric Field of a charged sphere with cylindrical gaussian surface
i could reconsider however it's something I'm being asked for. :(. yep i know the answer must be E=Q/(4pi*r^2)ε. and i know also the charge on the sphere it's all over the "surface area" and that the distance from the center to the point would be R+0.00000100 m. Once i got the Electric...- Molderish
- Post #3
- Forum: Advanced Physics Homework Help
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Electric Field of a charged sphere with cylindrical gaussian surface
So the problem statement is: A conducting solid sphere (R = 0.167 m, q = 6.63·10–6 C) is shown in the figure. Using Gauss’s Law and two different Gaussian surfaces, determine the electric field (magnitude and direction) at point A, which is 0.00000100 m outside the conducting sphere. (Hint: One...- Molderish
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- Charged Charged sphere Cylindrical Electric Electric field Field Gaussian Gaussian surface Sphere Surface
- Replies: 3
- Forum: Advanced Physics Homework Help
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Electric Field due to a charged line; With y as function of theta
Well the professor asked us for bring both methods in terms of y as function of theta & theta as function of y; I already did in terms of dy. making the substitution of cosine(θ) = x/√x^2+y^2 and then integrating. but now i need the other method. y in terms of "θ"- Molderish
- Post #3
- Forum: Introductory Physics Homework Help
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Electric Field due to a charged line; With y as function of theta
Well the main problem here its once a i get the integral i do not how to put dy as function of thetha and then integrate ; at the end that's what i get Ex=integral (dE*cos\theta)=(\lambda/4\pi\epsilon) times the integral of cos(\theta)dy/x^2+y^2 where i can easily find the cosine and...- Molderish
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- Charged Electric Electric field Field Function Line Theta
- Replies: 3
- Forum: Introductory Physics Homework Help
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Calculus III; standard form of & angle to eliminate the cross produc
Actually at first a tried with this result. but it said ,it was "incorrect" i guess because i didn't reorder the equation.- Molderish
- Post #11
- Forum: Calculus and Beyond Homework Help
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Calculus III; standard form of & angle to eliminate the cross produc
yeah actually i make the change of variables & i end up.. with "uv" terms. then i found the angle to make them 0. because it was the cross product terms then i just got a u^2 and v^2 with sin & cos =6 at this point i didn't know how to reorder the equation. that's why i go back to the...- Molderish
- Post #9
- Forum: Calculus and Beyond Homework Help
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Calculus III; standard form of & angle to eliminate the cross produc
oh ok! i see.. well i think is a minor detail i couldn't see , thanks so much!- Molderish
- Post #7
- Forum: Calculus and Beyond Homework Help
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Calculus III; standard form of & angle to eliminate the cross produc
but wouldn't it be x=ucos−vsin--> so ((ucos(pi/4)-vsin(pi/4))^2)/12 & y=usin+vcos----> ((usin(pi/4))+(vcos(pi/4)))^2/4?? did you reduce factors?- Molderish
- Post #5
- Forum: Calculus and Beyond Homework Help
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Calculus III; standard form of & angle to eliminate the cross produc
Because what i tried to do its to re-write the equation " x^2+xy+y^2=6" which i found easier. then i plugged into it the "x" and "y" the problems gives. but what i am not really sure if it means "elipse standard form"; i think the angle i got it's right since "...- Molderish
- Post #3
- Forum: Calculus and Beyond Homework Help
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Calculus III; standard form of & angle to eliminate the cross produc
Hi; I've been trying to solve the problem myself but i really don't what could be wrong; The problem says : Make the change of variables x=ucos−vsin y=usin+vcos where the angle 0<(phi)<2 is chosen in order to eliminate the cross product term in x^2+xy+y^2=6 Then find the standard form of...- Molderish
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- Angle Calculus Calculus iii Cross Form Standard
- Replies: 10
- Forum: Calculus and Beyond Homework Help