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Chapter 24. The Electric Field

The Electric Field Intensity
24-1. A charge of +2 (C placed at a point P in an electric field experiences a downward force of 8 x 10-4 N. What is the electric field intensity at that point?
[pic][pic][pic][pic]; E = 400 N/C, downward

24-2. A –5 nC charge is placed at point P in Problem 24-1. What are the magnitude and direction of the force on the–5 nC charge? (Direction of force F is opposite field E)
F = qE = (-5 x10-9 C)(-400 N/C); F = 2.00 x 10-6 N, upward

24-3. A charge of –3 (C placed at point A experiences a downward force of 6 x 10-5 N. What is the electric field intensity at point A?
A negative charge will experience a force opposite to the field.
Thus, if the –3 (C charge has a downward force, the E is upward.[pic][pic]; E = 20 N/C, upward

24-4. At a certain point, the electric field intensity is 40 N/C, due east. An unknown charge, receives a westward force of 5 x 10-5 N. What is the nature and magnitude of the charge?
If the force on the charge is opposite the field E, it must be a negative charge.
[pic] q = -1.25 (C

24-5. What are the magnitude and direction of the force thatwould act on an electron (q = -1.6 x 10-19 C) if it were placed at (a) point P in Problem 24-1? (b) point A in Problem 24-3?
The electric force on an electron will always be opposite the electric field.
(a) F = qE = (-1.6 x 10-19 C)(-400 N/C); F = 6.40 x 10-17 N, upward
(b) F = qE = (-1.6 x 10-19 C)(+20 N/C); F = -3.20 x 10-18 N, downward

24-6. What must be themagnitude and direction of the electric field intensity between two horizontal plates if one wants to produce an upward force of 6 x 10-4 N on a +60-(C charge? (The upward force on +q means E is also upward.)
[pic]; E = 10.0 N/C, up

24-7. The uniform electric field between two horizontal plates is 8 x 104 C. The top plate is positively charged and the lower plate has an equalnegative charge. What are the magnitude and direction of the electric force acting on an electron as it passes horizontally through the plates? (The electric field is from + to -, i.e., downward; force on e is up.)
F = qE = (-1.6 x 10-19 C)(8 x 104 N/C); F = 1.28 x 10-14 N, upward

24-8. Find the electric field intensity at a point P, located 6 mm to the left of an 8-(C charge. Whatare the magnitude and direction of the force on a –2-nC charge placed at point P?
[pic];
E = 2.00 x 109 N/C, toward Q
F = qE = (-2 x 10-9 C)(2.00 x 109 N/C)
F = -4.00 N, away from Q
24-9. Determine the electric field intensity at a point P, located 4 cm above a –12-(C charge. What are the magnitude and directionof the force on a +3-nC charge placed at point P?
Electric field will be downward, since that is the direction a positive charge would move.
[pic]; E = -6.75 x 107 N/C, downward
F = qE = (3 x 10-9 C)(-6.75 x 107 N/C) ; F = -0.202 N, downward

Calculating the Resultant Electric Field Intensity
24-10. Determine the electric filed intensity at the midpoint of a 70 mm line joininga –60-(C charge with a +40-(C charge.
[pic]
[pic]; ER = E1 + E2 (Both to left)
ER = -4.41 x 108 N/C – 2.94 x 108 N/C ; ER = 7.35 x 108 N/C. toward –60 (C

24-11. An 8-nC charge is located 80 mm to the right of a +4-nC charge. Determine the field intensity at the midpoint of a line joining the two charges.
[pic]
[pic]; ER = E1 + E2(E1 right, E2 left)
ER = -4.50 x 104 N/C + 2.25 x 104 N/C ; ER = -2.25 x 104 N/C, left
Note: The directions of the E field are based on how a test + charge would move.

24-12. Find the electric field intensity at a point 30 mm to the right of a 16-nC charge and 40 mm to the left of a 9-nC charge.
[pic]
[pic]; ER = E1 + E2 (E1 right, E2 left)...
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