A Point Charge in an Electric Field — Question 6

Question 6

A ramped field gives the same impulse as its mean but a different displacement

A particle starts from rest in a spatially uniform field $E_x(t)=E_0t/\tau$ for $0\le t\le\tau$. Its mass is m and charge is $q>0$. Other forces are negligible.

1. Integrate the acceleration to find velocity and displacement at the end of the pulse.

2. Compare with using the constant time-average field $E_0/2$. Identify which final quantity is reproduced and which is not.

Question 6 - Solution

A ramped field gives the same impulse as its mean but a different displacement

(a) Method and calculation

$$a_x(t)=\frac{qE_0}{m\tau}t,\quad v_x(t)=\frac{qE_0t^2}{2m\tau}.$$$$v_x(\tau)=\frac{qE_0\tau}{2m},\quad x(\tau)=\frac{qE_0\tau^2}{6m}.$$

(b) Interpretation and further calculation

$$\bar E=E_0/2:\quad v_{\mathrm{avg}}(\tau)=\frac{qE_0\tau}{2m},\quad x_{\mathrm{avg}}(\tau)=\frac{qE_0\tau^2}{4m}.$$$$x_{\mathrm{avg}}/x_{\mathrm{exact}}=3/2.$$Impulse depends on the time integral of force, whereas displacement also depends on when the acceleration occurs.

Physical check

Later acceleration contributes less time to displacement than earlier acceleration of the same impulse.