POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL

Question

I want help with this question about electrical energy and capacitance The question:- Find the change in electric potential energy associated with the electron as it goes on from x =0.120 m to x= - 0.180 m. I've found the electric potential energy it's about -7.20×10^-17J

The next part of this question is find by use conservation of energy the electron’s speed at x= -0.180m. The answer is 4.5% of the speed of light which is 1.35×10^7. I couldn't understand the answer of this question I considered the initial speed of the electron is zero then I got 1.25×10^7m/s which is not correct Please I want an idea about this problem

Answer 1

when dealing with particle physics of ions and electrons being accelerated in electromagnetic

fields it is customary to keep the units in eV's and use the relativistic equations for conservation

of energy:

E_total = E_kinetic + E_rest

In this equation:

E_kinetic is usually represented by T (just old school tradition for nuclear physicists)

E_rest = m_oc², the total energy when the particle is at rest

E_total = mc² , where m is the relativistic mass due to the particle's velocity relative to light

that is

m=m_o/√(1-v²/c²) , where v is the particle's velocity and c is the velocity of light

When you know the kinetic energy gained by the particle, you work the equations backwards,

so to speak, to figure out v.

let's do an example (stick with me here, the numbers will be very similar to yours in the end);

Let's say an electron is accelerated through an electric field from A to B, and the electric potential

from A to B, is -500V (+500V from B to A). Then the electron will gain a kinetic energy of 500eV

from A to B. Since an eV is the energy to move a charge of one electron through 1V, and the work

done is just qV.

... see how easy that is!

Just multiple the charge (in multiples of q, the charge on one electron) times the volts.

Now, if the electron has gain 500eV of kinetic energy, how fast is it traveling?

rearranging the above equation:

mc² = T + m_oc²

where

m₀c² = 0.511 MeV (a quantity to be found in any modern physics textbook tables)

so

mc² = 500eV + 0.511Mev = 0.511MeV + 0.0005MeV = 0.5115Mev

and

m_o/m = (m_oc²)/(mc²) = .511/.5115 = 0.99902248 = √(1-v²/c²)

thus

v²/c² = 1 - (0.99902248)² = 0.0019540786

v/c = 0.0442049

v = 0.0442049(c) = 0.0442049(3x10⁸m/s) = 1.32 x 10⁷ m/s answer

as you can see, in my example the electron goes thru -500V, is going 4.4% the speed of light

which is 1.32x10⁷m/s.

Your problem must be very similar, go thru it again while looking at my example.