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