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... To derive the moment of the couple, consider two vectors , rA and rB from O to points A and B lying on the line of action of F and - F. The moment of the couple about O is: M = rA x (F) + rB x (-F) = (rA - rB) x F By triangle law of vector addition, rB + r = rA or r = rA - rB So: M = r x F This resu ...
... To derive the moment of the couple, consider two vectors , rA and rB from O to points A and B lying on the line of action of F and - F. The moment of the couple about O is: M = rA x (F) + rB x (-F) = (rA - rB) x F By triangle law of vector addition, rB + r = rA or r = rA - rB So: M = r x F This resu ...
Chapter 22 -Gauss`s Law
... IV. Applications of Gauss’s Law Remember that Gauss’s Law can be used qualitatively to determine the location of charges and quantitatively to determine the strength of the electric field for relatively simple charge distributions. Remember also that in electrostatics, E = 0 inside a conductor. (Do ...
... IV. Applications of Gauss’s Law Remember that Gauss’s Law can be used qualitatively to determine the location of charges and quantitatively to determine the strength of the electric field for relatively simple charge distributions. Remember also that in electrostatics, E = 0 inside a conductor. (Do ...
Review for 16-17
... and midway between them is deflected upward by a potential of .0120 V. F. Through what potential was the electron accelerated to reach a velocity of 114,700 m/s from rest? Vq = 1/2mv2, q = 1.602x10-19 C, v = 114,700, m = 9.11x10-31 kg ...
... and midway between them is deflected upward by a potential of .0120 V. F. Through what potential was the electron accelerated to reach a velocity of 114,700 m/s from rest? Vq = 1/2mv2, q = 1.602x10-19 C, v = 114,700, m = 9.11x10-31 kg ...
Static-chapter3
... To derive the moment of the couple, consider two vectors , rA and rB from O to points A and B lying on the line of action of F and - F. The moment of the couple about O is: M = rA x (F) + rB x (-F) = (rA - rB) x F By triangle law of vector addition, rB + r = rA or r = rA - rB So: M = r x F This resu ...
... To derive the moment of the couple, consider two vectors , rA and rB from O to points A and B lying on the line of action of F and - F. The moment of the couple about O is: M = rA x (F) + rB x (-F) = (rA - rB) x F By triangle law of vector addition, rB + r = rA or r = rA - rB So: M = r x F This resu ...
Tutorial 4b - Electric Potential
... Tutorial 4b – Electric Potential 23–1 Electric Potential 1. (I) What potential difference is needed to stop an electron that has an initial velocity y 5.0 3 105 m/s? ...
... Tutorial 4b – Electric Potential 23–1 Electric Potential 1. (I) What potential difference is needed to stop an electron that has an initial velocity y 5.0 3 105 m/s? ...
Electric Field Mapping
... Figure 2: Upper side of the field mapping board 4) Fasten a sheet of paper to the upper side of the mapping board. Secure the paper by slipping it under the four rubber bumpers. (If you are using A4 papers, you may slide the paper under only the top two bumpers). 5) Select the right design template ...
... Figure 2: Upper side of the field mapping board 4) Fasten a sheet of paper to the upper side of the mapping board. Secure the paper by slipping it under the four rubber bumpers. (If you are using A4 papers, you may slide the paper under only the top two bumpers). 5) Select the right design template ...
DAY ONE - Rutgers Physics
... uniformly over the volume between the plates, so that the current density is just I/(2πR2 ). The problem is then equivalent to finding the magnetic field inside a fat wire of radius R. Using Ampere’s Law one finds B = µ0 rI/(2πR2 ) = µ0 rαt/(2πR2 ). The magnetic field lines form horizontal concentri ...
... uniformly over the volume between the plates, so that the current density is just I/(2πR2 ). The problem is then equivalent to finding the magnetic field inside a fat wire of radius R. Using Ampere’s Law one finds B = µ0 rI/(2πR2 ) = µ0 rαt/(2πR2 ). The magnetic field lines form horizontal concentri ...
Recitation ch 24
... In figure (5), a hollow sphere, of radius r that carries a negative charge -q, is put inside another hollow sphere, of radius R that carries a positive charge Q. At a distance x from the common center, such that r < x < R, the electric potential is: ...
... In figure (5), a hollow sphere, of radius r that carries a negative charge -q, is put inside another hollow sphere, of radius R that carries a positive charge Q. At a distance x from the common center, such that r < x < R, the electric potential is: ...
Electric field
... Coulomb’s Law and Vectors • The net force acting on an object is the sum of all the forces acting on it • For charged objects, these forces can be calculated using Coulomb’s Law and remembering that force vectors have direction • It is usually better to ignore signs and just deal with direction of f ...
... Coulomb’s Law and Vectors • The net force acting on an object is the sum of all the forces acting on it • For charged objects, these forces can be calculated using Coulomb’s Law and remembering that force vectors have direction • It is usually better to ignore signs and just deal with direction of f ...
Experiment 1G Uniform Circular Motion
... r = radius of the circle We can find an equation for the velocity v=∆x ∆t by calculating the distance and time for one revolution. The distance is the circumference of the circle 2r and the time is the period of revolution T. So: v = 2r T ...
... r = radius of the circle We can find an equation for the velocity v=∆x ∆t by calculating the distance and time for one revolution. The distance is the circumference of the circle 2r and the time is the period of revolution T. So: v = 2r T ...
CTChargesEFields
... Q-14. A metal sphere has a net charge +Q which is spread uniformly over its surface. How does the magnitude of the electric field at point A inside near the surface compare to the magnitude of the field at point B at the center of the sphere. A) EA > EB B) EA < EB ...
... Q-14. A metal sphere has a net charge +Q which is spread uniformly over its surface. How does the magnitude of the electric field at point A inside near the surface compare to the magnitude of the field at point B at the center of the sphere. A) EA > EB B) EA < EB ...
