Question

Two particles of charges and masses (+q${}_1$, m${}_1$) and (-q${}_2$, m${}_2$) are released 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 1

Let initial separation between the particles be $l$ and this remains constant during the motion. If relative separation remains constant then relative velocity should be zero; hence relative acceleration of the particles should also be zero.

$a_r=a_1-a_2=0$

or $\frac{q_{1}E-F_e}{m_1}=\frac{F_e-q_{2}E}{m_2}$

or $\frac{(q_1{m}_2+q_2{m}_1)E}{m_1{m}_2}=F_e[\frac{1}{m_1}+\frac{1}{m_2}]$

or $(q_1{m}_2+q_2{m}_1)E=F_e(m_1+m_2)=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{l^2}(m_1+m_2)$

After solving we get, $l=\frac{1}{2}\sqrt{\frac{q_1{q}_2(m_1+m_2)}{\pi {\epsilon }_{0}E(q_2{m}_1+q_1{m}_2)}}$