3. Consider a uniformly charged, perfectly spherical bowling ball of radius 15cm centered at some origin and carrying a charge density of 27uC/m3. a.What would be the flux through a closed, square cardboard box with dimensions 40cm x 40cm x 40cm also centered at the same origin (assume the sides of the box are parallel to the coordinate planes and remains that way throughout this problem)? 064700 09.10e+04 016800 O37500 043100 01.41e+05 b.Next, a 70cm long rod of uniform linear charge density -5uC/m is shoved perpendicularly through a side of the box and out the other side so that 12 cm of the rod remains sticking outside of the box on the one side and 18 cm of the rod sticks out the other side of the box. The bowling ball remains where it is. What would then be the flux through the entire box? O-2.74e+05 O-3.86e+05 O-1.59e+05 O-71300 O-1.83e+05 O-5.96e+05 c.Next, the rod is removed and the box is magically shrunk to 8cm x 8cm x 8cm dimensions so that it is now completely imbedded the bowling ball. Assume that the bowling ball remains where it is and remains completely intact. The newly shrunken box remains centered at the origin. What would now be the flux through the box? 02340 05090 01560 O1360 O3.30e+03 0609 d. What would be the flux through the newly shrunken box if it was then moved so that it was now centered at +2 cm on the x-axis? 0609 05090 03.30e+03 O2340 O1560 01360 e.Keeping the box and ball where they are (the ball centered at the origin and the box centered at x=2cm), if a point charge of -4 nC is placed at x= -6cm, what would then be the flux through the box? 0609 03.30e+03 O2340 O1560 O5090 01360 f.If a second charge is now placed at the origin and the flux through the box becomes 1700 Nm2/C, what was the size and sign of that charge (in nC)? 01.06 00.476 01.22 O2.58 O1.83 03.98 +
3. Consider a uniformly charged, perfectly spherical bowling ball of radius 15cm centered at some origin and carrying a charge density of 27uC/m3. a.What would be the flux through a closed, square cardboard box with dimensions 40cm x 40cm x 40cm also centered at the same origin (assume the sides of the box are parallel to the coordinate planes and remains that way throughout this problem)? 064700 09.10e+04 016800 O37500 043100 01.41e+05 b.Next, a 70cm long rod of uniform linear charge density -5uC/m is shoved perpendicularly through a side of the box and out the other side so that 12 cm of the rod remains sticking outside of the box on the one side and 18 cm of the rod sticks out the other side of the box. The bowling ball remains where it is. What would then be the flux through the entire box? O-2.74e+05 O-3.86e+05 O-1.59e+05 O-71300 O-1.83e+05 O-5.96e+05 c.Next, the rod is removed and the box is magically shrunk to 8cm x 8cm x 8cm dimensions so that it is now completely imbedded the bowling ball. Assume that the bowling ball remains where it is and remains completely intact. The newly shrunken box remains centered at the origin. What would now be the flux through the box? 02340 05090 01560 O1360 O3.30e+03 0609 d. What would be the flux through the newly shrunken box if it was then moved so that it was now centered at +2 cm on the x-axis? 0609 05090 03.30e+03 O2340 O1560 01360 e.Keeping the box and ball where they are (the ball centered at the origin and the box centered at x=2cm), if a point charge of -4 nC is placed at x= -6cm, what would then be the flux through the box? 0609 03.30e+03 O2340 O1560 O5090 01360 f.If a second charge is now placed at the origin and the flux through the box becomes 1700 Nm2/C, what was the size and sign of that charge (in nC)? 01.06 00.476 01.22 O2.58 O1.83 03.98 +
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter23: Continuous Charge Distributions And Gauss's Law
Section: Chapter Questions
Problem 9P: (a) Consider a uniformly charged, thin-walled, right circular cylindrical shell having total charge...
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Transcribed Image Text:3. Consider a uniformly charged, perfectly spherical bowling ball of radius 15cm centered at some origin and carrying a charge density of 27uC/m3.
a.What would be the flux through a closed, square cardboard box with dimensions 40cm x 40cm x 40cm also centered at the same origin (assume the sides of the box are parallel to the coordinate planes and remains that way throughout this problem)?
064700 09.10e+04 016800 O37500 043100 01.41e+05
b.Next, a 70cm long rod of uniform linear charge density -5uC/m is shoved perpendicularly through a side of the box and out the other side so that 12 cm of the rod remains sticking outside of the box on the one side and 18 cm of the rod sticks out the
other side of the box. The bowling ball remains where it is. What would then be the flux through the entire box?
O-2.74e+05 O-3.86e+05 O-1.59e+05 O-71300 O-1.83e+05 O-5.96e+05
c.Next, the rod is removed and the box is magically shrunk to 8cm x 8cm x 8cm dimensions so that it is now completely imbedded the bowling ball. Assume that the bowling ball remains where it is and remains completely intact. The newly shrunken box
remains centered at the origin. What would now be the flux through the box?
02340 05090 01560 O1360 O3.30e+03 0609
d. What would be the flux through the newly shrunken box if it was then moved so that it was now centered at +2 cm on the x-axis?
0609 05090 03.30e+03 O2340 O1560 01360
e.Keeping the box and ball where they are (the ball centered at the origin and the box centered at x=2cm), if a point charge of -4 nC is placed at x= -6cm, what would then be the flux through the box?
0609 03.30e+03 O2340 O1560 O5090 01360
f.If a second charge is now placed at the origin and the flux through the box becomes 1700 Nm2/C, what was the size and sign of that charge (in nC)?
01.06 00.476 01.22 O2.58 O1.83 03.98
+
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