Question

A particle of mass m and charge q is subjected to combined action of gravity and a uniform horizontal electric field of strength E. It is projected from ground with speed v in the vertical plane parallel to the field at an angle θ to the horizontal. What is the maximum distance the particle can travel horizontally before striking ground.

Answer 1

Dear student
$$\begin{gathered} \sin cetheelectricfieldisindirectionparalleltouwhichisangle\theta wrthorizontal \\ nowtakingtwocomponentsofelectricfield \\ {E}_x=E\cos \theta and{E}_y=E\sin \theta \\ nowY=u_{y}t+\frac{1}{2}{a}_y{t}^2 \\ 0=u\sin \theta t+\frac{1}{2}(\frac{qE}{m}-g)t^2 \\ t=0andt=\frac{2u\sin \theta }{(g-\frac{qE}{m})} \\ X=u_{y}t+\frac{1}{2}{a}_y{t}^2 \\ X=u\cos \theta \times \frac{2u\sin \theta }{(g-\frac{qE}{m})}+\frac{1}{2}\frac{qE}{m}\times \frac{4u^2{\sin }^2\theta }{{\left(g-\frac{qE}{m}\right)}^2} \\ X=\frac{u^2\sin 2\theta }{(g-\frac{qE}{m})}+\frac{qE}{m}\times \frac{2u^2{\sin }^2\theta }{{\left(g-\frac{qE}{m}\right)}^2} \\ X=\frac{u^2\sin 2\theta }{x}+c\frac{2u^2{\sin }^2\theta }{x^2}\{takenc=\frac{qE}{m}andx=(g-\frac{qE}{m}\left)\right\} \\ X=\frac{u^2}{x}(\sin 2\theta +c2{\sin }^2\theta ) \\ X=\frac{u^2}{x}(\sin 2\theta +c\frac{1-\cos 2\theta }{1}) \\ X=\frac{u^2}{x}(\sin 2\theta +c-c\cos 2\theta )or\frac{u^2}{x}(c+\{\sin 2\theta -c\cos 2\theta \left\}\right) \\ nowforXtobemaxima\{\sin 2\theta -c\cos 2\theta \}tobemaximaandweknowthatmaximaofA\sin \theta +B\cos \theta +C=\sqrt{A^2+B^2}+C=\sqrt{1+c^2}+c \\ so \\ X=\frac{u^2}{x}(c+\{\sin 2\theta -c\cos 2\theta \}=\frac{u^2}{x}\times \sqrt{1+c^2}+c=\frac{u^2}{(g-\frac{qE}{m})}\times \sqrt{1+\frac{q^2{E}^2}{m^2}}+\frac{qE}{m} \\ nowX=Rang{e}_{max}=\frac{u^2}{(g-\frac{qE}{m})}\times \sqrt{1+\frac{q^2{E}^2}{m^2}}+\frac{qE}{m} \\ Regards \end{gathered}$$