Answer:
The speed of proton when it emerges through the hole in the positive plate is [tex]2.05\times 10^5\ m/s[/tex].
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, [tex]u=3\times 10^5\ m/s[/tex]
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 :
[tex]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[/tex]
So, the speed of proton when it emerges through the hole in the positive plate is [tex]2.05\times 10^5\ m/s[/tex].
