Question

An electron is released from rest at the negative plate of a parallel-plate capacitor. If the distance across the plate is 2.0 mm and the potential difference across the plate is 6.0 V, with what velocity does the electron hit the positive plate? ( me = 9.1 × 10^−31 kg, e = 1.6 × 10^−19 C)

Answer 1

Answer:

Velocity of electron will be $1.325\times 10^6m/sec$

Explanation:

We have given distance across the plate d = 2 mm $=2\times 10^{-3}m$

Potential difference V = 6 volt

We know that potential difference at any distance is given by

V = Ed , here V is potential difference, E is electric field and d  is distance

So $E=\frac{V}{d}=\frac{6}{2\times 10^{-3}}=3000N/C$

Charge on electron $e=1.6\times 10^{-19}C$

We know that expression of velocity is given by $v=\sqrt{\frac{2qEd}{m_e}}$, here q is charge on electron, E is electric field and d is distance

So $v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 3000\times 3000}{9.11\times 10^{-31}}}=1.325\times 10^6m/sec$