exam1_solutions - Physics 222 Exam 1 Fall 2011 1 A) 116 cm...

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Unformatted text preview: Physics 222 Exam 1 Fall 2011 1 A) 116 cm to the right of the negative charge 11 E) 27 21 C) 90 2 B) C N / 10 9 . 3 4 12 C) 8.4 22 A) B 3 E) 13 C ) 30 V 23 D ) rotates 4 E) 14 B) conservation of charge 24 B ) 24 2 9.3 10 Am M- = 5 B) L Q k V 2 = 15 B) 2 - 2I 2- 4I 3 = 0 25 A) d qvI 2 , parallel to current 6 C) s m / 10 5 . 3 7 16 E) mA 5 . 7 A) Capacitance will increase 17 D) 2.9 V 8 D) 6000 N/C 18 B ) 0.16 T 9 A) 5.3 V 19 B) into the page 10 A) 850 W 20 D ) m 2 10 7 . 2- Physics 222 Exam 1 Fall 2011 1. A C 70 . 4 + and a C 50 . 3 - charge are placed 18.5 cm apart. Where can a third charge be placed so that it experiences no net force? A) 116 cm to the right of the negative charge B) 58 cm to the left of the negative charge C) 8 cm to the right of the negative charge D) 116 cm to the right of the positive charge E) 58 cm to the right of the positive charge Solution: Assume that the negative charge is d = 18.5 cm to the right of the positive charge, on the x-axis. To experience no net force, the third charge Q must be closer to the smaller magnitude charge (the negative charge). The third charge cannot be between the charges, because it would experience a force from each charge in the same direction, and so the net force could not be zero. And the third charge must be on the line joining the other two charges, so that the two forces on the third charge are along the same line. See the diagram. Equate the magnitudes of the two forces on the third charge, and solve for x > 0. ( 29 ( 29 ( 29 ( 29 ( 29 2 1 2 1 2 2 2 1 2 6 2 6 6 1 2 3.5 10 C 18.5cm 116cm 4.7 10 C 3.5 10 C Q Q Q Q Q k k x d x d x Q Q Q x d Q Q--- = = = +- = = =-- F F r r 2. Calculate the magnitude of electric field at one corner of a square 1.00 m on a side if the other three corners are occupied by C 10 25 . 2 6- charges. A) C N / 10 8 . 1 4 B) C N / 10 9 . 3 4 C) C N / 10 8 . 7 4 D) C N / 10 5 . 4 5 E) C N / 10 5 . 6 5 Solution: The field at the upper right corner of the square is the vector sum of the fields due to the other three charges. Let the variable d represent the 1.0 m length of a side of the square, and let the variable Q represent the charge at each of the three occupied corners. + x 4.7 C 3.5 C 1 Q 2 Q Q d + 4.7 C 3.5 C 1 Q 2 Q d 1 E r 3 E r 2 E r 1 Q 2 Q 3 Q d Physics 222 Exam 1 Fall 2011 1 1 1 2 2 o 2 2 2 2 2 2 2 3 3 1 2 2 , 2 2 cos45 , 2 2 4 4 0 , x y x y x y Q Q E k E k E d d Q Q Q Q E k E k k E k d d d d Q Q E k E E k d d = = = = = = = = = = Add the x and y components together to find the total electric field, noting that x y E E = . 1 2 3 2 2 2 2 2 1 4 4 x x x x y Q Q Q E E E E k k k E d d d = + + = + + = + = 2 2 2 2 2 1 1 2 2 4 2 x y Q Q E E E k k d d = + = + = + ( 29 ( 29 ( 29 6 9 2 2 4 2 2.25 10 C 1 8.988 10 N m C 2 3.87 10 N C 2 1.00m- = + = 3. Electrostatic field just outside a 3.50-cm-radius metal ball is C N 10 75 . 2 2 and points toward the ball. What is the field at 3.0 cm from the center of the ball?toward the ball....
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exam1_solutions - Physics 222 Exam 1 Fall 2011 1 A) 116 cm...

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