A particle is uncharged and is thrown vertically upward from ground level with a speed of 55 m/s. As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 13 m/s. Finally, the particle is given a negative charge −q. Ignoring air resistance, determine the speed (in m/s) with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the same maximum height h.

Question

A particle is uncharged and is thrown vertically upward from ground level with a speed of 55 m/s.  As a result, it attains a maximum height h.  The particle is then given a positive charge +q  and reaches the same maximum height h  when thrown vertically upward with a speed of   13 m/s. Finally, the particle is given a negative charge −q.  Ignoring air resistance, determine the speed (in m/s) with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the same maximum height  h.

Accepted Answer

In uniform electric in vertical direction if (+ve) charge feels extra acceleration in downward direction, then (–ve) charge will feel acceleration in upward direction.
$v_{u n c h \arg e d} = 5 \sqrt{5} m / \sec$
v = 0, h = height
$v^{2} - u^{2} = - 2 (g) h$
$= - 2 g h$
$u_{q +} = 13 m / \sec$
v = 0, h = h
$v^{2} - u^{2} = 2 (g + \frac{F_{E}}{m}) h$
$0 - {(13)}^{2} = - 2 (g + \frac{F_{E}}{m}) h$
Let $u_{q -} = u (s a y)$
v = 0 h = ht
$v^{2} - u^{2} = - 2 (g - \frac{F_{E}}{m}) h$
$- u^{2} = - 2 (g - \frac{F_{E}}{m}) h; u = 9 m / \sec$

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