A particle having charge same as that on an electron and mass 1.6 x 10^-30 kg is projected with an initial speed ‘u’ to the horizontal from the lower plate of a parallel plate capacitor as shown. The plates are sufficiently long and have separation 2cm. Then the maximum value of velocity of particle to not hit the upper plate is (E =10³V m upwards).

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$3 \times 10^{6} m / s$

$2 \sqrt{2} \times 10^{6} m / s$

$4 \times 10^{6} m / s$

$6 \times 10^{6} m / s$

answer is A.

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Detailed Solution

Maximum height, $h = \frac{u^{2} \sin^{2} θ}{2 (\frac{EQ}{m})}$
here, h = plate separation

$\Rightarrow 0.02 = \frac{u^{2} {(\frac{1}{\sqrt{2}})}^{2}}{2 (\frac{1.6 \times 10^{- 19} \times 10^{3}}{1.6 \times 10^{- 30}})} \Rightarrow u = 2 \sqrt{2} \times 10^{6} m / s$

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