Two particles of chargaes and masses `(+q_(1),m_(1))` and `(+q_(2),m_(2))` are reelaesed at two different points in a uniform electric field E established in free space. If their separation remains unchanged find the separation l between them .

Answer

`l=(1)/(2)sqrt((q_(1)q_(2)(m_(1)+m_(2)))/(pi epsilon_(0)E(q_(2)m_(1)+q_(1)m_(2)))`.

Explanation

Let initial separation between the particles be l and this remains constant during te motion. If relative separation remains constant then relative velocity should be zero, hence relative acceleration of the particle should also be zero.
,
`a_(r)=a_(1)-a_(2)=0`
or, `(q_(1)E-F_(e))/(m_(1))=(F_(e)-q_(2)E)/(m_(2))`
or `(q_(1)m_(2)+q_(2)m_(1))/(m_(1)m_(2))=F_(e)[(1)/(m_(1))+(1)/(m_(2))]`
or `(q_(1)m_(2)+q_(2)m_(1))E=F_(e)(m_(1)+m_(2))`
`=(1)/(4 pi epsilon_(0))(q_(1)q_(2))/(l^(2))(m_(1)+m_(2))`
After solving we get `l=(1)/(2)sqrt((q_(1)q_(2)(m_(1)+m_(2)))/(pi epsilon_(0)E(q_(2)m_(1)+q_(1)m_(2)))`.