- #1

jaskamiin

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Suppose we have an electric field, and push a proton or electron through it (from rest). We're accelerating it through an electric potential difference of some kind. Why do we apply conservation of energy to the particle-field system to find the speed of the particle?

Example:

Calculate the speed of a proton that is accelerated from rest through a ΔV of 150V.

Applying conservation of energy (and using kinetic energy?),

[tex]K_i + Ui = K_f + U_f[/tex]

becomes the following conservation of energy equation:

[tex]0 + qV_i = \frac{1}{2}m_pv_p^2 + 0[/tex]

Solving for the initial potential energy, we get [tex]qV_i = 2.4 \times 10^{-17}J [/tex]

Plug that back into the equation and solve for final velocity:

[tex]v_f = \sqrt{\frac{2qV_i}{m_p}} = \sqrt{\frac{2(2.4 \times 10^{-17} J)}{1.67 \times 10^{-27}kg }} = 1.695 \times 10^5 \frac{m}{s}[/tex]

Example:

Calculate the speed of a proton that is accelerated from rest through a ΔV of 150V.

Applying conservation of energy (and using kinetic energy?),

[tex]K_i + Ui = K_f + U_f[/tex]

becomes the following conservation of energy equation:

[tex]0 + qV_i = \frac{1}{2}m_pv_p^2 + 0[/tex]

Solving for the initial potential energy, we get [tex]qV_i = 2.4 \times 10^{-17}J [/tex]

Plug that back into the equation and solve for final velocity:

[tex]v_f = \sqrt{\frac{2qV_i}{m_p}} = \sqrt{\frac{2(2.4 \times 10^{-17} J)}{1.67 \times 10^{-27}kg }} = 1.695 \times 10^5 \frac{m}{s}[/tex]

- Why do we use this to find velocity? Simply because it contains the variables we need in order to calculate it?
- Why does U_i = qV_i ?

- Why does U_f = 0 ?

- Likewise when we accelerate an electron through the same potential difference, why do we set both initial kinetic and potential energies to 0 ?

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