Question

two charges 3microC and -5 microC are separated by distance 20cm . find the external dipole moment​

Answer 1

Answer:

This problem can be solved in a straightforward way or in a clever way. I’m going to present the straightforward way first, because I think it’s important to understand techniques for solving problems systematically. But the clever way is much easier.

As you know, the presence of the point charge qq will induce a redistribution of the charges on the surface of the conducting sphere. They will redistribute themselves such that the electric field is cancelled within the sphere. That is, the scalar potential ϕϕ is constant in the region r<Rr<R . Now, if we place the point charge on the positive z-axis, the system has azimuthal symmetry, and the scalar potential in the r<Rr<R region has an inner axial multipole expansion

ϕ(r,θ)=14πϵ0∑ℓ=0∞IℓrℓPℓ(cosθ)ϕ(r,θ)=14πϵ0∑ℓ=0∞IℓrℓPℓ(cos⁡θ)

Since the field vanishes in r<Rr<R , the potential is a constant. This implies that the interior multipole moments IℓIℓ vanish for all ℓ≥1ℓ≥1 . That is, the multipole moments of the induced charge on the surface of the conducting sphere must cancel that of the point charge at distance aa .

The interior multipole moments are given by

Iℓ=∫ρ(r′,θ′)r′ℓ+1Pℓ(cosθ′)d3r′Iℓ=∫ρ(r′,θ′)r′ℓ+1Pℓ(cos⁡θ′)d3r′

Since Pℓ(cos0)=Pℓ(1)=1Pℓ(cos⁡0)=Pℓ(1)=1 for all ℓℓ , the contribution to the multipole moment IℓIℓ made by the point charge, which has charge density qδ(x)δ(y)δ(z−a)qδ(x)δ(y)δ(z−a) , is q/aℓ+1q/aℓ+1 . For ℓ≥1ℓ≥1 , the interior multipole moment due to the charge on the sphere must be −q/aℓ+1−q/aℓ+1 . For ℓ=0ℓ=0 , the interior multipole moment due to the sphere is just the total surface charge QQ divided by the radius RR . So we have

I0Iℓ=Q/R=−qaℓ+1I0=Q/RIℓ=−qaℓ+1

where the second equation is for ℓ≥1ℓ≥1 and II now refers to the multipole moments due to the sphere alone.

To get the force on the point charge, we just need to calculate the field produced by the charge on the sphere. Since we want the field at r=ar=a , which is outside the sphere, we need to write the potential in the outer axial multipole expansion,

ϕ(r,θ)=14πϵ0∑ℓ=0∞Mℓrℓ+1Pℓ(cosθ)ϕ(r,θ)=14πϵ0∑ℓ=0∞Mℓrℓ+1Pℓ(cos⁡θ)

where I have used the letter M for the exterior axial multipole moments, which are defined by

Mℓ=∫r′ℓρ(r′,θ′)Pℓ(cosθ′)d3r′Mℓ=∫r′ℓρ(r′,θ′)Pℓ(cos⁡θ′)d3r′

Comparing the expression for the exterior multipole moments with those of the interior multipole moments, since the charge on the sphere is at a constant radius r=Rr=R , we find Mℓ=R2ℓ+1IℓMℓ=R2ℓ+1Iℓ , or

M0Mℓ=Q=−qR2ℓ+1aℓ+1M0=QMℓ=−qR2ℓ+1aℓ+1

We are specifically interested in the potential for θ=0θ=0 . We know by symmetry that the field will be directed radially, so we simply need to find ϕ(r,0)ϕ(r,0) and take its (negative) radial derivative to find the electric field affecting the point charge. Referring back to the expression for the potential in terms of the exterior multipole moments,

ϕ(r,0)=14πϵ0[Qr−q∑ℓ=1∞R2ℓ+1aℓ+1rℓ+1]ϕ(r,0)=14πϵ0[Qr−q∑ℓ=1∞R2ℓ+1aℓ+1rℓ+1]

Taking the radial derivative at r=ar=a and flipping the sign, we obtain

E(a,0)r=14πϵ0[Qa2−q∑ℓ=1∞(ℓ+1)R2ℓ+1a2ℓ+3]=14πϵ0a2[Q−q∑ℓ=1∞(ℓ+1)(R/a)2ℓ+1]E(a,0)r=14πϵ0[Qa2−q∑ℓ=1∞(ℓ+1)R2ℓ+1a2ℓ+3]=14πϵ0a2[Q−q∑ℓ=1∞(ℓ+1)(R/a)2ℓ+1]

This infinite arithmetico-geometric series may be explicitly summed, giving

E=14πϵ0(Qa2−qaR(a2−R2)2)E=14πϵ0(Qa2−qaR(a2−R2)2)

and the force on the point charge as F=qEF=qE .

The clever solution involves “guessing” that the potential outside the sphere is as though the induced charge on the sphere were concentrated at a point at radius R2/aR2/a with value −qR/a−qR/a , and arguing that since this potential function meets the boundary conditions, it must be the correct one for the region outside the sphere (not inside). See Method of image charges - Wikipedia or chapter 3 of Griffiths for more details; Griffiths explains it much better than I can.

Answer 2

Two charges , one +5uc and another -5uc are placed 1m apart. Calculate the dipole moment

Hope its help u