Wednesday, Aug. 31, 2005
... the direction. • Since the field permeates through the entire space, drawing vector arrows is not a good way of expressing the field. • Electric field lines are drawn to indicate the direction of the force due to the given field on a positive test charge. – Number of lines crossing unit area perpend ...
... the direction. • Since the field permeates through the entire space, drawing vector arrows is not a good way of expressing the field. • Electric field lines are drawn to indicate the direction of the force due to the given field on a positive test charge. – Number of lines crossing unit area perpend ...
ELECTRICITY I
... • How does the electric force between two charges change when the distance between them is doubled? • The force is quartered ...
... • How does the electric force between two charges change when the distance between them is doubled? • The force is quartered ...
Physics - Electrostatics Tutorial Question 1 – Fun with Tape a) Press
... Pip and Jed are playing a game of Pig-In Pig-Out. The players draw charged stones from a bag and place them on the board. The first player (Jed) tries to place stones so that the pig will be in an electric field, while the second player (Pip) tries to place stones so that there is no net field where ...
... Pip and Jed are playing a game of Pig-In Pig-Out. The players draw charged stones from a bag and place them on the board. The first player (Jed) tries to place stones so that the pig will be in an electric field, while the second player (Pip) tries to place stones so that there is no net field where ...
The Electric Field
... – show the direction of an electric field - away from positive and toward negative. – show the intensity of an electric field: • Lines bunched together field is stronger • Lines farther apart field is weaker ...
... – show the direction of an electric field - away from positive and toward negative. – show the intensity of an electric field: • Lines bunched together field is stronger • Lines farther apart field is weaker ...
2 nC
... Assess: Notice the N and C cancel out leaving units of m. Comparing with Problem 20.10, the answer of 2.6 cm seems to be in the right ballpark. ...
... Assess: Notice the N and C cancel out leaving units of m. Comparing with Problem 20.10, the answer of 2.6 cm seems to be in the right ballpark. ...
Magnetic Fields
... considering a straight segment of wire of length L and crosssectional area A, carrying a current I in a uniform magnetic field B, as shown in Figure. The magnetic force exerted on a charge q moving with a drift velocity vd is ...
... considering a straight segment of wire of length L and crosssectional area A, carrying a current I in a uniform magnetic field B, as shown in Figure. The magnetic force exerted on a charge q moving with a drift velocity vd is ...
Chapter 34
... u =uE + uB = εoE2 = B2 / μo When this is averaged over one or more cycles, the total average becomes uavg = εo(E2)avg = ½ εoE2max = B2max / 2μo ...
... u =uE + uB = εoE2 = B2 / μo When this is averaged over one or more cycles, the total average becomes uavg = εo(E2)avg = ½ εoE2max = B2max / 2μo ...
CHAPTER 23 SOLUTION FOR PROBLEM 19 (a) To calculate the
... the center of the shell, select A so the field does not depend on the distance. Use a Gaussian surface in the form of a sphere with radius rg , concentric with the spherical shell and within it (a < rg < b). Gauss’ law will be used to find the magnitude of the electric field a distance rg from the s ...
... the center of the shell, select A so the field does not depend on the distance. Use a Gaussian surface in the form of a sphere with radius rg , concentric with the spherical shell and within it (a < rg < b). Gauss’ law will be used to find the magnitude of the electric field a distance rg from the s ...
