An infinitely long line of charge has linear charge density 6.00×10−12 C/m . A proton (mass 1.67×10−27 kg, charge +1.60×10−19 C) is 12.0 cm from the line and moving directly toward the line at 4.10×103 m/s .a) Calculate the proton's initial kinetic energy. Express your answer with the appropriate units. b) How close does the proton get to the line of charge? Express your answer with the appropriate units.

a)        K = 1.4036 10⁻²⁰ J , b)   r = 1.23 m

Explanation:

a) The kinetic energy of the proton is

            K = ½ m v²

Let's calculate

          K = ½ 1.67 10⁻²⁷ (4.10 10³)²

           K = 1.4036 10⁻²⁰ J

b) At the point of closest approach the hundred and potential energy are equal

           U = K

           q E = K

To cellular the electric field let's use Gauss's law

             Ф = E dA = / ε₀

We define a Gaussian surface as a cylinder with a base perpendicular to the charge line, whereby the radius of the cylinder and the intensity of the electric field are parallel, whereby the scalar product is reduced to the algebraic product

          E A = q_{int} / ε₀

The cylinder area is

           A = 2π r l

We use the concept of linear density for the load inside

            λ = q_{int} / l

           q_{int} = λ l

We substitute

            E 2π r l = λ l /ε₀

             E = λ / 2π ε₀r

At the point of closest approach

            q E = k

            q λ / 2πε₀ r = K

            r = q λ / 2πε₀ K

Calculous

           r = 1.60 10⁻¹⁹ 6.00 10⁻¹² / (2π  8.85 10⁻¹² 1.4036 10⁻²⁰)

           r = 1.23 m

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer