Two insulating plates are both uniformly charged in such a way that potential difference between them is (V₂ -V₁ = 20 V) (i.e., plate 2 is at higher potential). The plates are separated by d = 0.1 m and can be treated as infintely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2 ?
(e = 1.6 x 10^-19 C, m_e = 9.11x 10^-31 kg)

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$7 .02 \times 10^{12} m / s$

$1.87 \times 10^{6} m / s$

$32 \times 10^{- 19} m / s$

$2 \cdot 65 \times 10^{6} m / s$

answer is A.

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Detailed Solution

Given that V₂ >V₁ i.e., the electric field will point, from plate 2 to plate 1 as shown in fig. The moving electron will experience au electric force, opposite to the direction of electric field.
Using work energy theorem, we have
$\frac{m_{e} v^{2}}{2} - 0 = e (V_{2} - V_{1})$
or $\frac{9 \cdot 11 \times 10^{- 31}}{2} v^{2} = 1 \cdot 6 \times 10^{- 19} \times 20$
$∴ V = \sqrt{(\frac{1 .6 \times 10^{- 19} \times 40}{9 .11 \times 10^{- 31}})} or V = 2 .65 \times 10^{6} m / s .$