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An-electric-dipole-is-placed-at-rest-in-a-uniform-external-electric-field-and-released-Discuss-its-motion-mathematically-




Question Number 40770 by Necxx last updated on 27/Jul/18
An electric dipole is placed at rest  in a uniform external electric  field,and released.Discuss its  motion mathematically.
$${An}\:{electric}\:{dipole}\:{is}\:{placed}\:{at}\:{rest} \\ $$$${in}\:{a}\:{uniform}\:{external}\:{electric} \\ $$$${field},{and}\:{released}.{Discuss}\:{its} \\ $$$${motion}\:{mathematically}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
let angle between axis of dipole and electric  field is θ  length og dipole =2l   charge on each pole are  +q  and −q   external electric field=E  so net force on dipole =qE+(−q)E=0  Torque with respect to centre of dipole is  ⌈^→ =l^→ ×qE^→ +(−l^→ )×(−qE^→ )     =2ql^→ ×E^→      ⌈=2qlEsinθ
$${let}\:{angle}\:{between}\:{axis}\:{of}\:{dipole}\:{and}\:{electric} \\ $$$${field}\:{is}\:\theta \\ $$$${length}\:{og}\:{dipole}\:=\mathrm{2}{l}\:\:\:{charge}\:{on}\:{each}\:{pole}\:{are} \\ $$$$+{q}\:\:{and}\:−{q}\:\:\:{external}\:{electric}\:{field}={E} \\ $$$${so}\:{net}\:{force}\:{on}\:{dipole}\:={qE}+\left(−{q}\right){E}=\mathrm{0} \\ $$$${Torque}\:{with}\:{respect}\:{to}\:{centre}\:{of}\:{dipole}\:{is} \\ $$$$\overset{\rightarrow} {\lceil}=\overset{\rightarrow} {{l}}×{q}\overset{\rightarrow} {{E}}+\left(−\overset{\rightarrow} {{l}}\right)×\left(−{q}\overset{\rightarrow} {{E}}\right) \\ $$$$\:\:\:=\mathrm{2}{q}\overset{\rightarrow} {{l}}×\overset{\rightarrow} {{E}}\: \\ $$$$\:\:\lceil=\mathrm{2}{qlEsin}\theta \\ $$

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