A long cylindrical shell of radius Rcarries a uniform surface charge on σ0the upper half and an opposite charge -σ0on the lower half (Fig. 3.40). Find the electric potential inside and outside the cylinder.

Short Answer

Expert verified

The expression for the potential inside the cylinder is2σ0Rπε0K=1,3,51k2sinkϕ(sR)k and outside the cylinder is 2σ0Rπε0K=1,3,51k2sinkϕ(Rs)k.

Step by step solution

01

Write the given data from the question.

The radius if the cylinder is R.

The uniform surface charge in upper half of the cylinder is σ0.

The uniform surface charge in upper lower of the cylinder is σ0.

02

Determine the formulas to calculate the electric potential inside and outside the cylinder.

outside the cylinder.

The expression for the potential is given as follows.

V(s,ϕ)=a0+b0lns+K=1α[spk(akcoskϕ+bksinkϕ)+s-k(ckcoskϕ+dksinkϕ)]…… (1)

Here, ak,bk, ckand dkare the constant.

From equation (1)

The potential inside the shell is given as follows.

Vin(s,ϕ)=K=1αsk(akcoskϕ+bksinkϕ) ……. (2)

The potential outside the shell is given as follows.

Vout(s,ϕ)=K=1αs-k(ckcoskϕ+dksinkϕ) …… (3)

03

Calculate the electric potential inside and outside the cylinder.

At the boundary condition, potential is continuous at .s=RHence, equate the potential inside and outside the cylinder.

K=1αRk(akcoskϕ+bksinkϕ)=K=1αR-k(ckcoskϕ+dksinkϕ)

Compare the coefficient ofcoskϕandsinkϕfrom the above equation.

akRk=Rkckck=R2kak

Similarly,

bkRk=Rkdkdk=R2kbk

Consider the equation which relates the normal derivative of the potential with the surface charge density.

Vs|R+Vs|R=σε0 ……. (4)

Calculate the derivative of potential inside cylinder.

Vs|R+=ss-k(ckcoskϕ+dksinkϕ)s=RVs|R+=(ksk+1)(ckcoskϕ+dksinkϕ)s=RVs|R+=(kRk+1)(ckcoskϕ+dksinkϕ)

Calculate the derivative of potential outside cylinder.

Vs|R=ssk(akcoskϕ+bksinkϕ)s=RVs|R=(ksk1)(akcoskϕ+bksinkϕ)s=RVs|R=(kRk1)(akcoskϕ+bksinkϕ)

Recall the equation (4),

Vs|R+Vs|R=σε0

Substitute (kRk+1)(ckcoskϕ+dksinkϕ)forVs|R+and(kRk1)(akcoskϕ+bksinkϕ)for Vs|Rinto above equation.

(kRk+1)(ckcoskϕ+dksinkϕ)+(kRk1)(akcoskϕ+bksinkϕ)=σε0

Substitute R2kakforckand R2kbkfordkinto above equation.

(kRk+1)(R2kakcoskϕ+R2kbksinkϕ)(kRk1)(akcoskϕ+bksinkϕ)=σε0(kRk1)(akcoskϕ+bksinkϕ)(kRk1)(akcoskϕ+bksinkϕ)=σε02kRk1(akcoskϕ+bksinkϕ)=σε0 ……(5)

Noe defines the above equation for the intervals,

2kRk1(akcoskϕ+bksinkϕ)={σ0ε0           (0<ϕ<π)σ0ε0         (π<ϕ<2π)

The value of integral of above angles,

02πsinkϕcoslϕdϕ=002πcoskϕcoslϕdϕ={0        klπ        k=l  

Multiply the equation (5) with coslϕ.

2kRk1[02πakcoskϕcoslϕdϕ+02πbksinkϕcoslϕdϕ]=σ0ε0[02πcoslϕdϕ02πcoslϕ]limδx02kRk1πak=σ0ε0[sinlϕl]0π2kRk1πak=σ0ε0(0)ak=0

The value ofakis becomes zero therefore multiply withsinlϕand integrate.

2kRk1[02πakcoskϕsinlϕdϕ+02πbksinkϕsinlϕdϕ]=σ0ε0[02πsinlϕdϕ02πcoslϕ]2kRk1πbk=σ0ε0[(coslϕl)0π+(coslϕl)02π]bk=σ0lε0(22coslπ)bk=σ02kπlε0Rk1(22coslπ)

The value ofπis obtained at the condition .k=l

bk={0                      if l is even2σ0πk2Rk1ε0           if l is odd

Similarly for the outside of the cylinder.

2kRk1[02πckcoskϕsinlϕdϕ+02πdksinkϕsinlϕdϕ]=σ0ε0[02πsinlϕdϕ02πcoslϕ]2kRk1πdk=σ0ε0[(coslϕl)0π+(coslϕl)02π]dk=σ0lε0(22coslπ)dk=σ02kπlε0Rk1(22coslπ)

The value of πis obtained at the condition .k=l

dk={0                      if l is even2σ0πk2Rk1ε0           if l is odd

Substitute 2σ0πk2Rk1ε0forbk and 0forakinto equation (2).

Vin(s,ϕ)=K=1αsk((0)coskϕ+2σ0πk2Rk1ε0sinkϕ)Vin(s,ϕ)=K=1αsk(2σ0πk2Rk1ε0sinkϕ)Vin(s,ϕ)=2σ0Rπε0K=1,3,51k2sinkϕ(sR)k

Substitute2σ0πk2Rk1ε0limδx0fordkand0forckinto equation (3).

Vout(s,ϕ)=K=1αs-k((0)coskϕ+2σ0πk2Rk1ε0sinkϕ)Vout(s,ϕ)=K=1αs-k(2σ0πk2Rk1ε0sinkϕ)Vout(s,ϕ)=2σ0Rπε0K=1,3,51k2sinkϕ(Rs)k

Hence, the expression for the potential inside the cylinder is and outside the cylinder is .

2σ0Rπε0K=1,3,51k2sinkϕ(sR)kand outside the cylinder is2σ0Rπε0K=1,3,51k2sinkϕ(Rs)k

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