A non-uniformly charged insulating sphere has a volume charge density p that is expressed asp= Brwhere B is a constant, and r is the radius from the center of the sphere. If the, the total charge of the sphere is Q and its maximum radius is R. What is the value for B?Sol.By definition, the volume charge density is expressed infinitesimally asp=where inis the infinitesimal charge andis the infinitesimal volume.So, we hayep = dq/So we can write this asdq = BdVBut,dV =drBy substitution, we get the followingdq = 4BTdrUsing Integration operation and evaluating its limits, the equation, leads toQ =BTRearranging, we getB =/( T

There are three different types of charge densities in general. The charge distributed per unit…

A non-uniformly charged insulating sphere has a volume charge density p that is expressed as p= Br where B is a constant, and r is the radius from the center of the sphere. If the, the total charge of the sphere is Q and its maximum radius is R. What is the value for B? Sol.

A non-uniformly charged insulating sphere has a volume charge density p that is expressed as p= Br where B is a constant, and r is the radius from the center of the sphere. If the, the total charge of the sphere is Q and its maximum radius is R. What is the value for B? Sol.

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Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

A non-uniformly charged insulating sphere has a volume charge density p that is expressed as
p= Br
where B is a constant, and r is the radius from the center of the sphere. If the, the total charge of the sphere is Q and its maximum radius is R. What is the value for B?
Sol.
By definition, the volume charge density is expressed infinitesimally as
p=
where in
is the infinitesimal charge and
is the infinitesimal volume.
So, we haye
p = dq/
So we can write this as
dq = B
dV
But,
dV =
dr
By substitution, we get the following
dq = 4BT
dr
Using Integration operation and evaluating its limits, the equation, leads to
Q =
BT
Rearranging, we get
B =
/( T

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