Question

A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative plate with a speed of 3.0 x 105m/s. What is its speed when it emerges through the hole in the positive plate? (Hint: The electric potential outside of a parallel-plate capacitor is zero).Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative plate with a m/s. What is its speed when it emerges through the hole in the positive plate? (Hint: The electric potential outside plate capacitor is

Answer 1

The speed of proton when it emerges through the hole in the positive plate is
`2.05* 10^5\ m/s`.

Step-by-step explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of,
`u=3* 10^5\ m/s`

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

`qV=(1)/(2)mv^2-(1)/(2)mu^2\\\\1.6*10^(-19)*250=(1)/(2)mv^2-(1)/(2)\cdot1.67*10^(-27)\cdot(3*10^(5))^(2)\\\\(1)/(2)mv^2=3.515\cdot10^(-17)\\\\v=\sqrt{(3.515\cdot10^(-17)\cdot2)/(1.67*10^(-27))}\\\\v=2.05* 10^5\ m/s`

So, the speed of proton when it emerges through the hole in the positive plate is
`2.05* 10^5\ m/s`.