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

Answer:

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

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, $u=3\times 10^5\ m/s$

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 :

$qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s$

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