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

(B) $2.65 \times 10^{6} \mathrm{~m} / \mathrm{s}$

(A) $32 \times 10^{-19} \mathrm{~m} / \mathrm{s}$

(C) $7.02 \times 10^{12} \mathrm{~m} / \mathrm{s}$

(D) $1.87 \times 10^{6} \mathrm{~m} / \mathrm{s}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Work, Power & Energy

Two insulating plates are both uniformly charged in such a way that the potential difference between them is

V_{2} - V_{1} = 20 V

. (i.e. plate 2 is at a higher potential). The plates are separated by

d = 0.1 m

and can be treated as infinitely 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 \times 1 0^{- 19} C

m_{e} = 9.11 \times 1 0^{- 31} kg

)

(A) $32 \times 1 0^{- 19} m / s$
(B) $2.65 \times 1 0^{6} m / s$ ✓ Correct
(C) $7.02 \times 1 0^{12} m / s$
(D) $1.87 \times 1 0^{6} m / s$

Answer: (B)

Solution

The electron is released from rest at the lower-potential plate and accelerated across to the higher-potential plate through a potential difference of

20 V

. The work done on it,

e V

, appears as kinetic energy:

\frac{1}{2} m v^{2} = e V

. Hence

v = \frac{2 e V}{m} = \frac{2 \times 1.6 \times 1 0 ^{- 19} \times 20}{9.11 \times 1 0 ^{- 31}} \approx 2.65 \times 1 0^{6} m/s

. The plate separation does not enter this energy balance.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Work, Power & Energy

Solution

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