A long cylindrical shell of radius Rcarries a uniform surface charge σ0on the 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.

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,ck and 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)forVs|Rinto above equation.

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

Substitute R2kakfor ckand 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]2kRk1πak=σ0ε0[sinlϕl]0π2kRk1πak=σ0ε0(0)ak=0

The value ofakis becomes zero therefore multiply with sinlϕand integrate.

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

The value ofπ is 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 ofπ is obtained at the condition k=l.

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

Substitute2σ0πk2Rk1ε0 for bkand 0for akinto 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

Substitute 2σ0πk2Rk1ε0for dkand0 forck into 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 is2σ0Rπε0K=1,3,51k2sinkϕ(sR)k and outside the cylinder is 2σ0Rπε0K=1,3,51k2sinkϕ(Rs)k.

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