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In the set-up shown, there is a large thin non-conducting vertical sheet with charge density $σ$ . In the nearby region a charge q of mass m is projected at an angle $θ$ with initial velocity $u$ . When this charge reaches its maximum height, find its velocity

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$u \sin θ (1 + \frac{σ q \tan θ}{m ε_{0} g})$

$u \sin θ (1 + \frac{σ q \tan θ}{2 m ε_{0} g})$

$u \cos θ (1 + \frac{σ q \tan θ}{m ε_{0} g})$

$u \cos θ (1 + \frac{σ q \tan θ}{2 m ε_{0} g})$

answer is A.

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Detailed Solution

$F_{y} = m g (- \hat{j})$ , $F_{x} = (\frac{σ}{2 ε_{0}}) q (\hat{i})$

$a_{y} = - g (\hat{i}) a_{x} = (\frac{σ q}{m 2 ε_{0}}) \hat{i}$

As $S_{y} = H_{\max}, V_{y} = 0$

Let the time taken to reach max. height be $t_{1}$

$v_{y} = u_{y} + a_{y} t_{1}$

$\Rightarrow 0 = u \sin θ - g t_{1} \Rightarrow t_{1} = \frac{u \sin θ}{g}$

At $t = t_{1}, v_{x} = u_{x} + a_{x} t_{1}$

$\Rightarrow v_{x} = u \cos θ + (\frac{σ q}{2 m ε_{0}}) \frac{u \sin θ}{g} = u \sin θ [1 + \frac{σ q \tan θ}{2 m ε_{0} g}]$

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