- #1
GingerBread27
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A charge Q = 1.95e-04 C is distributed uniformly along a rod of length 2L, extending from y = -14.2 cm to y = +14.2 cm, as shown in the diagram on your assignment. A charge q = 2.45e-06 C, and the same sign as Q, is placed at (D,0), where D = 60.0 cm.
Consider the situation as described above and the following statements. If the statement is true, answer T, if it is false, answer F, and if the answer cannot be determined from the information provided, answer C. For example if B and C are true and there is not enough information to answer D and the rest are false, then answer FTTCF.
A)The magnitude of the force on charge q due to the small segment dy is dF=(kqQ/16L*Lr*r)dy
B)The total force on q is generally in the east direction.
C)The net force on q in the x-direction equals zero.
D)The charge on a segment of the rod of infinitesimal length dy is given by dQ=(Q/4L*L)dy
E)The net force on q in the y-direction does equal zero.
Now I keep getting this wrong and I don't know why. As far as answers the only one I'm pretty sure about is that E is true because by symmetry the y components will cancel out. I put A is false because I'm not sure where the 16 is coming from. I put B as uncertain because you really can't be sure where the force is going and for D I put false because I get that dQ=Q/2L. We just started covering this material so I'm very lost.
A second question asks to : Use integration to compute the total force on q in the x-direction (in N). I know I haveto take the integral from one of the rod to the other but not sure what equation to integrate. Please help.
Consider the situation as described above and the following statements. If the statement is true, answer T, if it is false, answer F, and if the answer cannot be determined from the information provided, answer C. For example if B and C are true and there is not enough information to answer D and the rest are false, then answer FTTCF.
A)The magnitude of the force on charge q due to the small segment dy is dF=(kqQ/16L*Lr*r)dy
B)The total force on q is generally in the east direction.
C)The net force on q in the x-direction equals zero.
D)The charge on a segment of the rod of infinitesimal length dy is given by dQ=(Q/4L*L)dy
E)The net force on q in the y-direction does equal zero.
Now I keep getting this wrong and I don't know why. As far as answers the only one I'm pretty sure about is that E is true because by symmetry the y components will cancel out. I put A is false because I'm not sure where the 16 is coming from. I put B as uncertain because you really can't be sure where the force is going and for D I put false because I get that dQ=Q/2L. We just started covering this material so I'm very lost.
A second question asks to : Use integration to compute the total force on q in the x-direction (in N). I know I haveto take the integral from one of the rod to the other but not sure what equation to integrate. Please help.
