Recent content by origamipro
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Conceptual: two positive point charges value of E
Ok i see it now. The two charges exert destructive interference, but the waves don't disappear. Once Q1 wave passes Q2 it combines and it diminishes as 1/r^2.- origamipro
- Post #3
- Forum: Introductory Physics Homework Help
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Conceptual: two positive point charges value of E
Homework Statement Both Q1 = Q2 = +5E-6 and are harnessed to screen and in a vacuum.----Q1---A----Q2------B--------P this is x axis, A and B are distances Homework Equations kQ/r^2 The Attempt at a Solution So the E field at point P is KQ1/(A+B)^2 + KQ2/(B)^2 My question is Why...- origamipro
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- Charges Conceptual Point Point charges Positive Value
- Replies: 2
- Forum: Introductory Physics Homework Help
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Graduate What Determines the Value of km in Magnetic Field Calculations?
For a moving charge is km = μ0/2pi or is km = μ0/4pi? examples: Biot-Savart (magnetic field dB at a point P due to a length element ds that carrues a steady current I is: dB = (μ0/4pi) (I ds x r) / r^2 While for an long straight wire, (≥10m?) B = (μ0 I)/(2pi r) What is the different...- origamipro
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- Constant Field Magnetic Magnetic field
- Replies: 1
- Forum: Electromagnetism
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Conducting shell and proton with KE
Right that is what i showed. Ke = 1/2 m v^2 the integration is the delta V(voltage) = q(kQ1/3R - KQ2/R)- origamipro
- Post #8
- Forum: Introductory Physics Homework Help
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Conducting shell and proton with KE
Q1 = 4E-16 Q2 = ? q = 1.6E-19 qm = 1.67E-27 given answer Q2 = -3.96E-16 at the surface of the shell kinetic energy must equal zero. The potential energy at the surface does not need to be accounted for since it only wants value of Q2 when proton gets to the surface. Is this assumption...- origamipro
- Post #6
- Forum: Introductory Physics Homework Help
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Conducting shell and proton with KE
Bump. Q net is +.04E-16C. Both charge sources are modeled from the very center. How does the proton come to rest?- origamipro
- Post #4
- Forum: Introductory Physics Homework Help
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What is the flux through each face of the pyramid?
right on thanks. height is sqrt(3^2+4^2) Area of face of pyramid is 1/2 base x height = (.5)(6)(5) = 15- origamipro
- Post #3
- Forum: Introductory Physics Homework Help
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What is the flux through each face of the pyramid?
Homework Statement base of pyramid is a 6m x 6m square height of pyramid is 4meters. a 52 N/C E field orthogonal to the base of the pyramid Homework Equations EA cos theta The Attempt at a Solution if flux through base = EAcos theta = 52 (36m^2) cos(0) = 1872 N m^2 / C Area...- origamipro
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- Flux Pyramid
- Replies: 2
- Forum: Introductory Physics Homework Help
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Conducting shell and proton with KE
i have the option to right click the picture and open in a new tab, through my chrome browser. [IMG=http://img543.imageshack.us/img543/8674/15jd.jpg][/PLAIN] Uploaded with ImageShack.us- origamipro
- Post #3
- Forum: Introductory Physics Homework Help
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Conducting shell and proton with KE
Homework Statement http://imageshack.us/photo/my-images/543/15jd.jpg/ [IMG=[PLAIN]http://img543.imageshack.us/img543/8674/15jd.jpg][/PLAIN] Homework Equations Qnet outside sphere = Q1+Q2 KEi + PEi = KEf + PEf q = 1.6E-19 mass of q = 1.672E-27 The Attempt at a Solution +(q ∫ KQ1 / R^2 ]...- origamipro
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- Conducting Proton Shell
- Replies: 8
- Forum: Introductory Physics Homework Help
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Work on charge from outside to inside of sphere
After integration Kq/R + KQ/R^3 (.4704) KQ/R ( 1 + (1/R^2) (.4704)) KQ/R (11/8) q = answer- origamipro
- Post #3
- Forum: Introductory Physics Homework Help
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Find Applied Force: Mass, Accel Given
-F -3(9.8) + T = 3a 4.5(9.8) - T = 4.5a -F -3(9.8) +4.5(9.8) = 7.5a solve for F the force. 12.825 Newtons- origamipro
- Post #6
- Forum: Introductory Physics Homework Help
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Find Applied Force: Mass, Accel Given
4.5(a) = 4.5(9.8) - 3(9.8) - F if a = .25 F = 13.575 Newtons added to mass 2 in order to slow the decent of mass 1 to .25m/s^2. Assuming frictionless pulley.- origamipro
- Post #2
- Forum: Introductory Physics Homework Help
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Work on charge from outside to inside of sphere
Homework Statement A solid non-conducting sphere of radius R = 1.12m. The sphere posses a total charge Qtot spread uniformally throughout its volume. a) derive equations for electric field for 1) 0<r<R 2) r>R result in terms of r R and Q b) Derive an equation that gives...- origamipro
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- Charge Outside Sphere Work
- Replies: 3
- Forum: Introductory Physics Homework Help