An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How much work is required to bring an additional electron from infinity to the origin? Express your answer with the appropriate units. Part B If, instead of the second electron coming in from infinity, it is initially at x = + 20.00 m on the axis and is given an initial velocity of 100.0 m/s toward the origin, does it reach the origin? Part C How close to the origin does it come?

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aristocles aristocles · Answer 1 · 2019-02-28T02:08:52+01:00

PART A)

Electrostatic potential at the position of origin is given by

[tex]V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}[/tex]

here we have

[tex]q_1 = 1.6 \times 10^{-19} C[/tex]

[tex]q_2 = -1.6 \times 10^{-19} C[/tex]

[tex]r_1 = r_2 = 1 m[/tex]

now we have

[tex]V = \frac{Ke}{r} - \frac{Ke}{r}[/tex]

[tex]V = 0[/tex]

Now work done to move another charge from infinite to origin is given by

[tex]W = q(V_f - V_i)[/tex]

here we will have

[tex]W = e(0 - 0) = 0[/tex]

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

[tex]U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}[/tex]

[tex]U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}[/tex]

[tex]U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})[/tex]

[tex]U = 1.15\times 10^{-30}[/tex]

Now we know

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2[/tex]

[tex]KE = 4.55 \times 10^{-27} kg[/tex]

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin