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An ideal dipole of dipole moment p is held in front of an infinite grounded conducting plane as shown. The force on the dipole in equal to : (k=1/4π∈0)Page_4-1.svg

Asked by aanand10000 27th August 2017, 10:59 PM
Answered by Expert
Answer:
1) Our first step is to consider the method of images and assume an image dipole to be present along negative y-axis.
 
 
2) The distance between the conductor and dipole is r. So, the distance between dipole and its image will be 2r.
 
3) In order to find force, we need to find the interaction energy between the dipole and its image.
 
4) Interaction energy between them is
 
begin mathsize 12px style straight U equals negative straight P with rightwards harpoon with barb upwards on top times straight E with rightwards harpoon with barb upwards on top equals open parentheses negative straight P straight j with hat on top close parentheses times open parentheses fraction numerator 2 Pj over denominator 4 πε subscript 0 open parentheses 2 straight r close parentheses cubed end fraction close parentheses equals negative fraction numerator 2 kP squared over denominator open parentheses 2 straight r close parentheses cubed end fraction end style
 
5) Hence, the force on the dipole is
 
begin mathsize 12px style straight F equals negative fraction numerator partial differential straight U over denominator partial differential straight r end fraction equals fraction numerator negative 3 kP squared over denominator 4 straight r to the power of 4 end fraction end style
Answered by Expert 28th August 2017, 12:12 PM
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