Question

A particle having negative charge equal to that of an electron and mass $1.6\times {10}^{-30}kg$ is projected with an initial speed ${}^{\prime }{u}^{\prime }$ at an angle ${45}^o$ to the horizontal from the lower plate of a parallel plate arrangement as shown in figure. The plates are sufficiently long and have a separation of $cm$. Then the maximum value of the velocity of the particle with which it can be projected so that it does not hit the upper plate is :(Take the electric field between the plates as ${10}^{3}V{m}^{-1}$ directed upward and gravity is ignored)

• A. $2\times {10}^{6}m/s$
• B. ${10}^{6}m/s$
• C. $2\sqrt{2}\times {10}^{6}m/s$
• D. $\sqrt{2}\times {10}^{6}m/s$

Answer 1

The correct option is C $2\sqrt{2}\times {10}^{6}m/s$

$H=\frac{4^2{\sin }^2\theta }{2(g+\frac{qE}{m}}$
$4^2=\frac{{10}^{-2}\times 2\times }{\frac{1}{2}}[10+\frac{1.6\times {10}^{-13}\times {10}^3}{1.6\times {10}^{-30}}]$
$4^2=8\times {10}^{-2}[10+{10}^{14}]$
$4=2\sqrt{2}\times {10}^{-6}$