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Standard XII

Physics

Capacitance of Parallel Plate Capacitor

Two insulatin...

Question

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 to plate 2?
$e = 1.6 \times 10^{- 19} C, m_{e} = 9.11 \times 10^{- 31} k g$

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Solution

$\frac{1}{2} m v^{2} = e V$
$v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 20}{9.11 \times 10^{- 31}}} = 2.65 \times 10^{6} m / s$

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12mv2=eV v=√2eVm=√2×1.6×10−19×209.11×10−31=2.65×106m/s

$\frac{1}{2} m v^{2} = e V$
$v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 20}{9.11 \times 10^{- 31}}} = 2.65 \times 10^{6} m / s$