Given is:-
Radius of the sphere
total positive charge
Now,
The volume charge density of the sphere is
by plugging all the values we get
which gives us
Now,
We know that
substituting and all the other values we get
eq-1
this is for the electric field inside the sphere
Therefore
part -a
For r=0
plugging this value in eq-1 we get
Part-b
For r= 8.29 cm
which gives us
Part-c
r=40cm
For this Again using Gauss's law we get
by plugging all the values we get
which gives us
part-d
for r=42 cm
which gives us
id sphere of radius 10.0cm has a tctal positivc cnarg2 of 21,4 uc uniformly distributed throughout...
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