Find the energy stored in a uniformly charged solid sphere of radiusRand charge q.Do it three different ways:

(a)Use Eq. 2.43. You found the potential in Prob. 2.21.

(b)Use Eq. 2.45. Don't forget to integrate over all space.

(c)Use Eq. 2.44. Take a spherical volume of radiusa.What happens as a?

Short Answer

Expert verified

(a)The total energy stored in the uniformly charged sphere is W=14πε034q2R.

(b)The total energy stored in the uniformly charged sphere is W=14πε034q2R.

(c)The total energy stored in the uniformly charged sphere is W=14πε034q2R.

The contribution due to the surface integral becomes small as the valueaapproaches zero.

Step by step solution

01

Determine the expression for the energy stored.

Write the expression for the amount of energy stored in a uniformly charged sphere of radius Rand charge q.

W=12pVdτ......(1)

Here,σis the volume charge density, W is the stored energy V and is the potential.

02

Solve for part (a) determine the charge density.

(a)

The charge per unit volume of the sphere is defined as its volume charge density. It can be expressed in the following way:

p=qVolume

Here, qis the charge of the solid sphere.

The cube of the radius of the volume determines the volume of the sphere.

Volume=43πR3

Here, Ris the radius of the solid sphere.

Now, substitute 43πR3for Volumeof the sphere in the equation p=qVolume and Solving for the P.

p=q43πR3=3q4πR3


Therefore, the charge density is p=3q4πR3.

03

 Step 3: Determine stored energy.

(b)

Using the result of problem 2.21, a charged sphere of radius has the following potential:

V=14πε0qR32-r22R2

Write the expression for change in the volume of the sphere is,

dτ=ooxo2xrdrdθd=4πr2dr

Substitute 14πε0qR32-r22R2 for V , 3q4πR2for Pand4πr2drfordτin equationW=12pVdτ and solving for W.

W=3q216πε0R60R(3R2-r2)r2dr=3q216πε0R63R2r33-r550R=14πε035q2R

Thus, the total energy stored in the uniformly charged sphere is W=14πε035q2R

04

Determine the stored energy in the uniformly charged sphere.

(c)

Using the equation 2.45, determine the amount of energy contained in an evenly charged solid sphere.

Write the expression for the energy stored in the sphere is,

W=ε02E2dτ

Here, Eis the electric field intensity and is the permittivity of the free space.

The electric field of a conducting sphere is expressed as follows:localid="1654671404051" E=14πε0qrR3,r<R(insidethesphere)14πε0qr3,r>R(outsidethesphere)

Now, Substitute the electric field values inside and outsides the sphere and4πr2drfordτin the equationW=ε02E2dτand soling for W.localid="1654671414121" W=ε020Rqr4πε0R324πr2dr+ε020qr4πε0R324πr2dr=q28πε0R60Rr2dr+q28πε00Rr-2dr

Simplifying the above equation,

localid="1654671482761" W=q28πε0R6R55+q28πε01R=14πε0q2215R+1R=14πε035q2R

Thus, the total energy stored in the uniformly charged sphere is localid="1654671492285" W=14πε035q2R.

Using the equation 2.44, determine the amount of energy stored in an evenly charged solid sphere.

localid="1654671499692" w=ε02vE2dτ+sVE.da

Here, Wis the large enough to enclose all the charge, Eis the electric field, V is the Voltage and da is the small area.

The electric field of a conducting sphere is expressed as follows:

localid="1654671510116" E=14πε0qrR3,r<R(insidethesphere)14πε0qr2,r>R(outsidethesphere)

Let's use a radius of sphere a > R.

Substitute the expression for electric field for (r < R and r > R) in the equationlocalid="1654671519764" W=ε02VE2dτ+sVE.daand simplifying.

Solve as further,

W=q240πε0R+q28πε0R=14πε034q2R

Hence, the total energy stored in the uniformly charged sphere is W=14πε035q2R

The contribution due to the surface integral becomes small as the valueaapproaches zero.

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