An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to twice the initial value?

The distance is

Explanation:

From the question we are told that

         The initial speed of the  electron is

         The mass of electron is

         Let be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

         Let be the initial kinetic energy of the electron \

          Let be the kinetic energy of the electron at the distance from the proton

  Considering that energy is conserved,

  The energy at the initial position of the electron = The energy at the final position of the electron

      i.e

are the potential energy at the initial  position of the electron and at distance d of the electron to the proton

                Here

So the equation becomes

Here are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

  is electrostatic constant with value

i.e

            is the velocity at distance d from the proton = 2

  So the equation becomes

Making d the subject of the formula

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

b.

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