The graph below shows variation of electric potential in a newly developed compo

Question

The graph below shows variation of electric potential in a newly developed composite with distance along x-axis. (a) A positive charge Q equal to 4 q is released from rest at 6.00 nm. What is the x-component of the electric field at this point? (b) What is its potential energy at entry point? (c) What is the potential energy of Q when its kinetic energy (in ev) is maximum? (d) What is the maximum kinetic energy (in eV) of charge Q? (e) What is the turning point of Q? (f) If the particle were to reach region (bc), what would its initial total energy have to be at x = 6nm?

Explanation / Answer

a) charge = 4qe = 6.4e-19 C

The potential from h to g is varying uniformly

E = V/r = -6V//2nm    = -3.0 e+9 V/m

b) The potential at entry point 6 nm = -3V

Potential energy of the charge U = -3.0* 6.4e-19 = -19.2e-19 eV at entry point ( 6nm)

c) The charge would move to the left as the PE is lower and gain KE

The PE is minimum between g and h and KE is maximum

   PE = -7.5 * 6.4e-19 = -48e-19 eV

d)    KE = -19.2e-19 + 48.0e-19   = 28.8 eV gained.

e) at e the PE =0 , at entry point PE= -19.2e-19 ev

       KE gained = 19.2 eV

it will move forwad until it looses the whole KE and gain PE

at mid point between e and d i.e. 1 nm from the origin V = +3.0 V

and PE = 19.2e-19 eV

the charge would turn back at this point.

f) PE at bc = 12*6.4e-19 = 76.8e-19 eV

   PE at entry (x=6nm) = -19.2 eV

The charge need to have initial KE equal to the difference in PE

     = (76.8 +19.2) e-19 = 96.0e-19 eV

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