LAB2_ElectricField - PHYS-1200\/1250 Lab Manual Name Delano Munoz Whatts Section_1 21C \u2013 PhET Activity \u2013 Electric Field The electric force on a

LAB2_ElectricField - PHYS-1200/1250 Lab Manual Name...

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PHYS-1200/1250 Lab Manual Name: Delano Munoz Whatts_______ Section____1____21C – PhET Activity – Electric Field Download and run the PhET simulation, “Charges and Fields”. Click/Check “Electric Field”, “Values”, and “Grid” in upper right box. Drag a +1 nC charge to a grid point near the center of the grid. The arrows that appear indicate the direction of the field at the points at the center of the arrows.Drag the sensor tool (from the box at the bottom of the screen) around the grid and observe how the direction and magnitude change with position. You can drag many sensors to various locations to make it easier to visualize the vector field.1)What do you think the intensity of the field vectors indicates? object and an electric field produced by all other charges. In symbols, ´F0=q0´Eallother charges, and in words, “The force on particle 0 due to all other charges is equal to the charge q0multiplied by the field due to all other charges.” The force on charge q0due to pointcharge q1is ´F1on0=q0q1r102´r10|´r10|, so the field at the position of charge 0 due to point charge q1is ´E0=1q0kq0q1r102´r10|´r10|=kq1r102´r10|´r10|. Eq. 21.bThe field points away from a positive charge and decreases in strength as the inverse square of the distance from the source charge. The field at a point is the vector sum of the fields from all charges (except the test probe charge) in the problem. 2)Drag a field sensor (bottom toolbox) to distances of 1, 2, and 3 meters from the 1 nC charge. What is field strength at each of these points (Note that 1V/m = 1N/C.)? Compare your simulation result withthe field strength computed directly from equation 21b.1The electric force on a charged object can be thought of as resulting from an interaction between that object and an electric field produced by all other charges. In symbols, ´F0=q0´Eallother charges, and in words, “The force on particle 0 due to all other charges is equal to the charge q0multiplied by the field due to all other charges.” The force on charge q0due to pointcharge q1is ´F1on0=q0q1r102´r10|´r10|, so the field at the position of charge 0 due to point charge q1is ´E0=1q0kq0q1r102´r10|´r10|=kq1r102´r10|´r10|. Eq. 21.bThe field points away from a positive charge and decreases in strength as the inverse square of the distance from the source charge. The field at a point is the vector sum of the fields from all charges (except the test probe charge) in the problem. k
PHYS-1200/1250 Lab Manual Name: Delano Munoz Whatts_______ Section____1____3)Sketch field lines to represent thefield in the space to the right.

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