A Point Charge in an Electric Field — Question 5

Question 5

Integrating a nonuniform field avoids treating acceleration as constant

A positive particle moves along x from 0 to L in the static field $E_x(x)=E_0(1+x/L)$, with $E_0,L>0$. Its initial speed is $v_0$ and its mass and charge are m and q. Ignore other forces.

1. Use the work-energy theorem to find its speed at L.

2. Compare with approximating the field everywhere by its initial value $E_0$. Find the error in the predicted kinetic-energy gain.

Question 5 - Solution

Integrating a nonuniform field avoids treating acceleration as constant

(a) Method and calculation

$$W=q\int_0^LE_0(1+x/L)\,dx=\tfrac32qE_0L.$$$$\tfrac12m(v_L^2-v_0^2)=\tfrac32qE_0L\Rightarrow v_L=\sqrt{v_0^2+3qE_0L/m}.$$

(b) Interpretation and further calculation

$$\Delta K_{\mathrm{const}}=qE_0L,\quad\Delta K_{\mathrm{exact}}=\tfrac32qE_0L.$$$$\frac{\Delta K_{\mathrm{exact}}-\Delta K_{\mathrm{const}}}{\Delta K_{\mathrm{exact}}}=1/3.$$The average field along this straight path is $3E_0/2$, not its value at the start.

Physical check

The work integral remains valid even though the acceleration changes continuously with position.