GSO002 Problem 20
Problem Statement
Optical Tweezer Manipulation of a Polar Nanoparticle: Interaction with Mirror Dipole of Grounded Conducting Sphere
Problem Statement
In vacuum, a perfectly grounded conducting sphere of radius a is placed with its center at the origin O(0,0,0).
At a point A(0,0,r0) on the z-axis at distance r0 (r0>a) from the sphere, a polar nanoparticle of negligible size is fixed. This particle possesses a permanent electric dipole moment of magnitude p (referred to simply as the "dipole").
Initially, the dipole direction points toward the origin O (i.e., in the −z direction).
While the particle's center of mass position is kept at point A, an external manipulation (such as an optical tweezer) quasi-statically rotates the dipole until it is finally parallel to the positive x-axis direction.
Find the work W done by the external force on the particle during this rotation.
The spatial extent of the dipole itself is sufficiently small compared to r0−a, so it can be treated as an ideal point dipole. Let the proportionality constant in Coulomb's law be k. Gravity and all other interactions are neglected.
Constraints
- Coulomb's law constant: k=9.0×109 N⋅m2/C2
- Radius of conducting sphere: a=1.0×10−8 m
- Position of particle: r0=2.0×10−8 m
- Magnitude of dipole moment: p=3.0×10−25 C⋅m
Input Format
The work W [J] is expressed as W=X×10−18. Give the value of positive integer X.
Solution
1. Image Dipole Derivation
The image of a point charge q at z=r0 outside a grounded conducting sphere is a charge q′=−qa/r0 at position z′=a2/r0.
For dipole p=(px,0,pz) at A, considering each component:
Radial (z) component:
- Image point charge: qim=pzr02a
- Image dipole z-component: pim,z=pzr03a3
Tangential (x) component:
- Image dipole x-component: pim,x=−pxr03a3
All images are located at z′=a2/r0.
2. Electric Field at Point A from the Image
Let d=r0−r0a2=r0r02−a2.
x-component (equatorial field of image x-dipole):
Eim,x=k((r02−a2)/r0)3pxa3/r03=kpx(r02−a2)3a3z-component (axial field of image point charge and z-dipole):
Eim,z=kpz(r02−a2)3a(r02+a2)3. Electrostatic Interaction Energy
Since the electric field is induced by the dipole itself (via the conducting sphere), the interaction energy includes a factor of 21:
U(θ)=−21p⋅Eim=−2(r02−a2)3kp2a[a2+r02cos2θ]where θ is the angle between the dipole and the z-axis.
4. Work Done by External Force
W=Ufinal−Uinit- Initial state (θ=π, −z direction): Uinit=−2(r02−a2)3kp2a(a2+r02)
- Final state (θ=π/2, +x direction): Ufinal=−2(r02−a2)3kp2a(a2)
5. Numerical Calculation
- r02−a2=(4.0−1.0)×10−16=3.0×10−16
- Denominator: 2(3.0×10−16)3=54×10−48
- Numerator: (9.0×109)×(9.0×10−50)×(1.0×10−8)×(4.0×10−16)=324×10−65
X=60.
Answer: 60