only question 54 Thanks Link http://books.google.com/books?id=ecYWAAAAQBAJ&pg=PA
ID: 2273356 • Letter: O
Question
only question 54
Thanks
Link
http://books.google.com/books?id=ecYWAAAAQBAJ&pg=PA744&dq=find+the+charge+contained+within+a+sphere+of+radius&hl=en&sa=X&ei=HmbuUpIHtKmwBIy8gMgE&ved=0CDIQ6AEwAQ#v=onepage&q=find%20the%20charge%20contained%20within%20a%20sphere%20of%20radius&f=false
Explanation / Answer
a)
q = Q (4/3 pi r^3)/(4/3 pi a^3)
==> q = (r/a)^3 Q
b)
E = k q/r^2
==> E = k ((r/a)^3 Q)/r^2
==> E = k Q r/a^3
c)
Q
d)
E = k Q/r^2
e)
E = 0
(inside a conductor, electric field is zero)
f)
total charge must be zer:
q_b = -Q
g)
Q
h)
the spherical surface of radii "a"
(because it has the minimum surface)
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