Problem: Electric Field and Potential for a ChargedSphere An insulating sphere o

Question

Problem: Electric Field and Potential for a ChargedSphere
An insulating sphere of radius R has non-uniform charge density given by(1)

(r) = 0(1 r/R)

.

a) Determine the total charge contained in the sphere.

b) Using Gauss’s law, nd the electric eld as a function of r for r > R
(outside the sphere).

c) Now, using Gauss’s law again, nd the electric eld as a function of r for r < R (inside the sphere). Compare your answers to (b) and (c) at thepoint r = R.

d) Briey explain why it is okay to say the potential can be put to zero whenr . With this as the starting point, determine the electric potential asa function of r both outside and inside the sphere. (How should you handlethe point r = R?)

Explanation / Answer

part a:

total charge=integration of charge density*4*pi*r^2*dr

from r=0 to r=R

=integration of pho_0*(1-(r/R))*4*pi*r^2*dr

=4*pi*pho_0*((r^3/3)-(r^4/(4*R))

using limits from r=0 to r=R, we get

total charge=4*pi*pho_0*((R^3/3)-(R^3/4))

=pi*pho_0*R^3/3

part b:

outside the sphere, total charge enclosed=total charge on the sphere

=pi*pho_0*R^3/3

if electric field at a distance r is E,

then epsilon*E*4*pi*r^2=total charge enclosed (gauss law)

==>E=pho_0*R^3/(12*epsilon*r^2)

part c:

for r <R:

charge enclosed=4*pi*pho_0*((r^3/3)-(r^4/(4*R))

then using gauss' law,

epsilon*E*4*pi*r^2=4*pi*pho_0*((r^3/3)-(r^4/(4*R))

==>E=(pho_0/epsilon)*((r/3)-(r^2/(4*R))

at r=R, E=pho_0*R/(12*epsilon)

which is same as what obtained from part b with r=R

part d:

for a large distance r,

the net charge can be treated a point charge.

potential due to a point charge Q =k*Q/r

where k=coloumb's constant

when r--> infinity, 1/r --> 0

so potential at r --> infinity is 0.

for r>R :

E=pho_0*R^3/(12*epsilon*r^2)

potential V=integration of -E.dr

=(pho_0*R^3/(12*epsilon*r))+c

where c is a constant

at r=infinity, V=0

hence c=0

so V for r>R is given by (pho_0*R^3/(12*epsilon*r))

at r=R, potential =pho_0*R^2/(12*epsilon)

for r<R:

E=(pho_0/epsilon)*((r/3)-(r^2/(4*R))

then V=integration of (-E.dr)

=(pho_0/epsilon)*((r^3/(12*R))-(r^2/6))+c...(1)

where c is a constant

at r=R, V=pho_0*R^2/(12*epsilon)

from equation 1,

at r=R, V=-(pho_0/epsilon)*(R^2/12)+c

==>pho_0*R^2/(12*epsilon)=-(pho_0/epsilon)*(R^2/12)+c

==>c=(pho_0/epsilon)*(R^2/6)

hence for r<R, potential=-(pho_0/epsilon)*(R^2/12)+(pho_0/epsilon)*(R^2/6)

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