- #1

Physicslearner500039

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- Homework Statement
- A positron (charge +e, mass equal to the electron mass) is moving at 1.0 X 107 m/s in the positive direction of an x axis when, at x = 0, it encounters an electric field directed along the x axis. The

electric potential V associated with the field is given in Fig. 24-52. The scale of the vertical axis is set by Vs = 500.0 V. (a) Does the positron emerge from the field at x = 0 (which means its motion is reversed) or at x = 0.50 m (which means its motion is not reversed)? (b) What is its speed when it emerges?

- Relevant Equations
- NA

My attempt, pictorially it looks like

I am confused with the questions (a). Does the positron emerge from the field at x =0? There is no potential at x=0, so the positron will continue with the same speed hence its motion is not reversed.

For the (b). The maximum volt is 200V, if i apply the Ui+Ki = Uf+Kf;

0.5*mp*v_ini^2 = 500*e + 0.5*mp*v_fin^2; mp -> mass of proton, v_init = 1*10^7m/s

0.5*1.67*10^-27 * 10^14 = (500*1.69*10^-19) + (0.5*1.67*10^-27*V_fin^2)

Solving for V_fin = 9.9*10^33 m/s; The positron continues with this speed. Even the constant potential of 500V is applied continuously from 0.2m to 0.5m why the speed is not decreasing? Please advise.