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? a)4530 b)1210 c)2930 d)1390 e)542 f)2080   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 -3 nC is placed at x= -5cm, what would then be the flux through the box? a)2080 b)2930 c)4530 d)1210 e)1390 f)542   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)? a)1.08 b)4.14 c)5.82 d)2.76 e)2.40 f)8.99   I need answers for part d, e and f. Please answer the following parts neatly, correctly, completely and show all the steps. i will highly appreciate it!

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Consider a uniformly charged, perfectly spherical bowling ball of radius 15cm centered at some origin and carrying a charge density of 24uC/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)?
a)38300 b)80900 c)14900 d)57500 e)1.25e+05 f)33300

 

b.Next, a 70cm long rod of uniform linear charge density -1.5uC/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?
a)-29500 b)-62200 c)-96100 d)-11500 e)-44200 f)-25600

 

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?
a)2930 b)2080 c)1210 d)542 e)1390 f)4530

 

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?
a)4530 b)1210 c)2930 d)1390 e)542 f)2080

 

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 -3 nC is placed at x= -5cm, what would then be the flux through the box?
a)2080 b)2930 c)4530 d)1210 e)1390 f)542

 

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)?
a)1.08 b)4.14 c)5.82 d)2.76 e)2.40 f)8.99
 
I need answers for part d, e and f. Please answer the following parts neatly, correctly, completely and show all the steps. i will highly appreciate it!
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