N electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed ve, while the proton acquires a speed vp. Find the ratio ve/vp.

ve/vp = 42.8

Explanation:

According to the exercise, the proton accelerates from A to B. According to the principle of energy conservation:

Final energy of the proton at point B = KEp1B + EPEp1B

Total proton energy at point A = KEp1A + EPEp1A

KEp1B = EPEp1A - EPEp1B

The initial energy of the proton is equal to zero, since it is at rest. Therefore the electric potential of a charge will be equal to:

EPE = qv, where v is the charge potential

The proton's final kinetic energy will be equal to:

KEp1B = EPEp1A - EPEp1B = q (vA - vB) (eq. 1)

If an electron is negatively charged, we have that qe = -qp

The kinetic energy of the electron will be equal to:

KEp1A = EPEe1B - EPEe1A = -q (vB - vA) = q (vA - vB) (eq. 2)

Comparing equations 1 and 2:

KEp1B = KEe1A, (mp * vp ^ 2) ^ 2 = (me * ve ^ 2) / 2

We solve and solve for ve / vp:

ve / vp = (mp / me) ^ 1/2 = (1.67x10 ^ -27kg / 9.11x10 ^ -31kg) ^ 1/2 = 42.8

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