Uniformly Charged Rod A charge Q = 8.4 x 10-4 C is distributed uniformly along a rod of length L = 18.5 cm, extending from y=-L/2 to y=+L/2, as shown in the diagram below. A charge q = 6.85 x 10-6 C, and th Q, is placed at (D, 0), where D = 12 cm. Conceptual ForceCalculation dy Use integration to compute the total force on q in the a direction. fx= -0.0014062 N O L/2 Consider the situation as described above and the following statements. Check each box that corresponds to a true statement. Select "None of the above" if none of them are true. Q A. The charge on a segment of the rod of infinitesimal length dy is given by dQ = dy B. Given that both charges are positive, the net force on q in the y direction is upwards. C. The total force on q is generally in the + direction. D. Given that both charges are positive, the net force on q in the a direction is to the left. kqQ E. The magnitude of the force on charge q due to the small segment dy is d.F = dy F. The y-components of the force on q cancel. G. None of the above. -L/2

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Uniformly Charged Rod
A charge Q = 8.4 × 10−4 C is distributed uniformly along a rod of length L = 18.5 cm, extending from y = -L/2 to y = +L/2, as shown in the diagram below. A charge q = 6.85 × 10-6 C, and the same sign a
Q, is placed at (D, 0), where D = 12 cm.
Conceptual
ForceCalculation
dy
Use integration to compute the total force on q in the x direction.
O
fx = -0.0014062 N
y
L/2
Consider the situation as described above and the following statements. Check each box that corresponds to a true statement. Select "None of the above" if none of them are true.
-dy
L
A. The charge on a segment of the rod of infinitesimal length dy is given by dQ
B. Given that both charges are positive, the net force on q in the y direction is upwards.
C. The total force on q is generally in the + direction.
D. Given that both charges are positive, the net force on q in the direction is to the left.
kqQ
E. The magnitude of the force on charge q due to the small segment dy is dF = -dy
4Lr²
F. The y-components of the force on q cancel.
G. None of the above.
-L/2
r
Transcribed Image Text:Uniformly Charged Rod A charge Q = 8.4 × 10−4 C is distributed uniformly along a rod of length L = 18.5 cm, extending from y = -L/2 to y = +L/2, as shown in the diagram below. A charge q = 6.85 × 10-6 C, and the same sign a Q, is placed at (D, 0), where D = 12 cm. Conceptual ForceCalculation dy Use integration to compute the total force on q in the x direction. O fx = -0.0014062 N y L/2 Consider the situation as described above and the following statements. Check each box that corresponds to a true statement. Select "None of the above" if none of them are true. -dy L A. The charge on a segment of the rod of infinitesimal length dy is given by dQ B. Given that both charges are positive, the net force on q in the y direction is upwards. C. The total force on q is generally in the + direction. D. Given that both charges are positive, the net force on q in the direction is to the left. kqQ E. The magnitude of the force on charge q due to the small segment dy is dF = -dy 4Lr² F. The y-components of the force on q cancel. G. None of the above. -L/2 r
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