Question

A proton with an initial speed of 900,000 m/s is brought to rest by an electric field.

What was the potential difference that stopped the proton?
Express your answer to two significant figures and include the appropriate units.

Answer 1

To stop the proton, a potential difference of 290 V is required.

Step-by-step explanation:

To calculate the potential difference required to stop the proton, we need to use the equation:
KE = qV

Where KE is the kinetic energy, q is the charge of the proton (-1.60 x 10-19 C), and V is the potential difference. We know the initial kinetic energy of the proton is given by:
KE = (1/2)mv2

By equating the two expressions for kinetic energy, we can solve for V:
qV = (1/2)mv2

Plugging in the values for the charge (q = -1.60 x 10-19 C) and the mass of the proton (m = 1.67 x 10-27 kg), as well as the initial velocity of the proton (v = 900,000 m/s), we can solve for V:
V = (1/2)(m)(v2)/(q)

Calculating the value, we find that the potential difference required to stop the proton is approximately 290 V.