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Using the formulas of Chapter 12, Section 5, sum the series in (8.20) to get (8.21).

Short Answer

Expert verified

The sum series is derived given in (8.20).

V=qr22arcos+a2(Ra)qr2+(R2a)22r(R2a)cos

Step by step solution

01

Given Information:

It has been asked to use the formula given in section 5 of chapter 12.

02

Step 2:Uses of the Green function:

The green function is a function that is used to solve a corresponding partial differential equation in three dimensions, namely Poisson鈥檚 equation.

2u=f(r)=f(x,y,z)

03

Use the formula given in section 5 of chapter 12:

The summing the series in equation (8.20) from the book:

V=qr22arcos+a2qlR2l+1rl1Pl(cos)al+1

Modify the equation to bring the equation mentioned below.

V=qr22arcos+a2(Ra)qr2+(R2a)22r(R2a)cos

(In every sum in this problem, goes from 0 to ).

Calculate the sum using the equations mentioned below.

V=KR[12rRcos+(rR)2]1/2

V=KRlrlPl(cos)Rl=KRl(rR)lPl(cos)

04

Prove both the equation to be equal.

Rewrite the term with the sum in the equation.

qlR2l+1rl1Pl(cos)al+1=qRarl(R2ar)lPl(cos) 鈥︹赌.(1)

The factor (R2ar)in the sum in the last equation plays the same role as the factor (rR)in the sum of equation (5.17). Thus, comparing (5.17) with (5.12) and seeing where the factor (rR)is present inside the brackets of (5.12), It can be concluded what will be the result of the sum from equation (1) - it is equal to the bracket in (5.12) where (rR)change with (R2ar):

qlR2l+1rl1Pl(cos)al+1=qRar[12(R2ar)cos+(R2ar)2]1/2=(Ra)qr2+(R2a)22r(R2a)cos

Hence, it has been proved that summing the series in (8.20) gives us (8.21).

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Most popular questions from this chapter

Question:Let V=0in the Schrodinger equation (3.22) and separate variables in 2-dimensional rectangular coordinates. Solve the problem of a particle in a 2-dimensional square box, 0<x<l,0<y<lThis means to find solutions of the Schrodinger equation which are 0 for x=0,x=I,y=0,y=I, that is, on the boundary of the box, and to find the corresponding energy eigenvalues. Comments: If we extend the idea of a 鈥減article in a box鈥 (see Section 3, Example 3) to two or three dimensions, the box in 2D might be a square (as in this problem) or a circle (Problem 8); in 3D it might be a cube (Problem 7.17) or a sphere (Problem 7.19). In all cases, the mathematical problem is to find solutions of the Schrodinger equation with V=0inside the box and =0on the boundary of the box, and to find the corresponding energy eigenvalues. In quantum mechanics, describes a particle trapped inside the box and the energy eigenvalues are the possible values of the energy of the particle.

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