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what is the speed of the proton when it is accelerated from rest through a potential difference of -700 v ?

User Fiddlest
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The speed of a proton accelerated through a potential difference of -700 V is approximately 2.6 × 10^7 m/s.

The speed of a proton when it is accelerated from rest through a potential difference of -700 V can be calculated using the equation:

v = √(2qV/m)

Where v is the speed of the proton, q is the charge of the proton (1.6 × 10^-19 C), V is the potential difference (-700 V), and m is the mass of the proton (1.67 × 10^-27 kg). Plugging in the values, we get:

v = √(2 × 1.6 × 10^-19 C × -700 V / 1.67 × 10^-27 kg)

After performing the calculation, the speed of the proton is approximately 2.6 × 10^7 m/s.

User Giovane
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