A practical of mass ‘m’ carries an electric positive charge Q and is subjected to the combined action of gravity g and a uniform horizontal electric field of strength E. It is projected with speed u in the vertical plane parallel to the electric field and at an angle $θ$ to the horizontal. The maximum distance the particle travels horizontally when it is at the same level as its starting points is

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$\frac{{EQu}^{2}}{{mg}^{2}}$

$\frac{u^{2} (EQ + mg)}{{mg}^{2}}$

$\frac{u^{2}}{{mg}^{2}} (EQ + \sqrt{(mg)^{2} + (EQ)^{2}})$

$\frac{u^{2}}{g}$

answer is C.

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Detailed Solution

$\begin{array}{l} x = ucos θ + \frac{1}{2} \frac{EQ}{m} t^{2} \\ y = usin θt - \frac{1}{2} {gt}^{2} \\ \Rightarrow T = \frac{2 usin θ}{g} \\ \Rightarrow R = \frac{u^{2}}{g} \sin 2 θ + \frac{1}{2} \frac{QE}{m} \cdot \frac{4 u^{2} \sin^{2} θ}{g^{2}} \end{array}$

$\begin{array}{l} \frac{dR}{dθ} = 0 \Rightarrow \tan 2 θ = \frac{- mg}{EQ} \\ \Rightarrow R_{max} = \frac{u^{2}}{{mg}^{2}} (EQ + \sqrt{(mg)^{2} + (EQ)^{2}}) \end{array}$