Question

Two protons each of mass m and charge e are held at rest at distance 1 m apart in air. The maximum velocity acquired by each proton after release is​

Answer 1

v = e√(2k/m)

Explanation:

work done= change in kinetic energy

{work done = eV}

{ ∆K.E. = 1/2m(v^2 - u^2) }

As starting from rest u =0

1/2 m v^2 = eV

(V= kq/r)

v^2 = (2 e k e) / (m r)

( r= 1... given)

v^2 = (2 k e^2)/ m

v = e× √(2k/m)

✌️

Answer 2

Given:

Two protons each of mass m and charge e are held at rest at distance 1 m apart in air.

To find:

Max Velocity acquired by each proton after release?

Calculation:

Each of the proton will reach maximum velocity when the initial electrostatic potential energy will be fully converted to kinetic energy.

$PE= KE$

$\implies \: \dfrac{k(e)(e)}{1} = \dfrac{1}{2} m {( v_{max})}^{2}$

$\implies \: \dfrac{k {e}^{2} }{1} = \dfrac{1}{2} m {( v_{max})}^{2}$

$\implies \: m {( v_{max})}^{2} = 2k {e}^{2}$

$\implies \: { v_{max}}^{2} = \dfrac{2k {e}^{2} }{m}$

$\implies \: v_{max} = \sqrt{ \dfrac{2k {e}^{2} }{m}}$

$\implies \: v_{max} = e\sqrt{ \dfrac{2k }{m}}$

So, final answer is OPTION A)