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Concept Question: 5 Equal Charges Six equal positive charges q sit at the vertices of a regular hexagon with sides of length R. We remove the bottom charge. The electric field at the center of the hexagon (point P) is: 2kq 1. E = 2 ĵ R kq 3. E = 2 ĵ R 2kq 2. E = - 2 ĵ R kq 4. E = - 2 ĵ R 5. E = 0 1 Concept Question Answer: 5 Equal Charges Answer 4. kq E=- 2 R ĵ E fields of the side pairs cancel (symmetry) E at center due only to top charge (R away) Field points downward Alternatively: “Added negative charge” at bottom R away, pulls field down 2 Concept Question: Field Lines Electric field lines show: 1. Directions of forces that exist in space at all times. 2. Directions in which positive charges on those lines will accelerate. 3. Paths that charges will follow. 4. More than one of the above. 5. I don’t know. 3 Concept Question Answer: Field Lines Answer: 2. Directions positive charges accelerate. NOTE: This is different than flow lines (3). Particles do NOT move along field lines. (1) is not correct because we don’t know sign of charge so we don’t know direction of force 4 Concept Question: Electric Field Two charged objects are placed on a line as shown below. The magnitude of the negative charge on the right is greater than the magnitude of the positive charge on the left, qR > qL . Other than at infinity, where is the electric field zero? 1. 2. 3. 4. 5. Between the two charged objects. To the right of the charged object on the right. To the left of the charged object on the left. The electric field is nowhere zero. Not enough info – need to know which is positive. 5 Concept Question Answer: Electric Field Answer: 3. To the left of the positively charged object on the left http://web.mit.edu/viz/EM/visualizations/electrostatics/ElectricFieldConfigurations/pcharges/pcharges.htm Between: field points from positively charged object to negatively charged object On right: field is dominated by negatively charged object On left: electric field due to close smaller positively charged object will cancel electric field due to further larger negatively charged object. 6