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Imagine that a proton released at rest from point A moves in response to the electric field of a fixed charge distribution along a path that takes it through point B. If the potential has a value of 3500 V at A and -1500 V at B, what is the proton's speed as it passes point B

Answer :

The speed of the proton is [tex]9.79*10^5m/s[/tex]

Data;

  • potential difference (A) = 3500V
  • potential difference (B) = -1500V

Velocity of The Proton

The work done to move through a potential velocity 'v' is q

The potential difference 'v' is the difference between A and B

[tex]V = 3500-(-1500) = 5000v[/tex]

But the work is converted into kinetic energy of proton.

[tex]q_pV = \frac{1}{2}m_pV^2\\[/tex]

Let's substitute the values and solve for the velocity

[tex](1.6*10^-^1^9)*(5000)=\frac{1}{2}(1.67*10^-^2^7)v^2\\v^2 = \frac{2* 1.6*10^-^1^9 * 5000}{1.67*10^-^2^7}\\ v = \sqrt{9.58*10^1^1} \\v = 9.79 * 10^5m/s[/tex]

The speed of the proton is [tex]9.79*10^5m/s[/tex]

Learn more about work done on an electron here;

https://brainly.com/question/13673636

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