7.1.27 | revision #14664 | Savchenko Solutions

Statement

7.1.27.

a. The electron enters an axisymmetric electric field created by stationary
charges and initially moves parallel to the field axis at a distance $$r$$ from it.
Electron velocity $$v$$ . If the velocity of the electron and the distance from it
to the axis change slightly when moving in the field, then the momentum
acquired by the electron can be estimated by the formula $$p_⊥$$ = $\frac{eq}{2πε0vr}$
, where
q is the total electric charge inside a cylindrical region of radius r. Use the
Gauss theorem to derive this formula.

b. Determine the transverse momentum acquired by charge $$q_1$$ , which flew
past charge $$q_2$$ . The minimum distance between the charges r, the charge
velocity $$q_1$$ was initially equal to $$v$$ and changed slightly.

c. Estimate the minimum distance from the nucleus of a nitrogen atom at
which an electron accelerated by a potential difference of $100 \hspace{0.1cm} (kV)$ flew, if it was
deflected by the nucleus by an angle of $10^{\text -3} \hspace{0.1cm} (rad)$ .

Solution

(a)

Since the electric field is axisymmetric, we will use superposition, where an electron interacts with a point charge q.

$E(\theta) = \frac{kq}{(r/sin(\theta))^2}$
$F(\theta)=eE(\theta)=\frac{kq}{(r/sin(\theta))^2}$
$$$dr=vdt$$$
$F_y=Fsin(\theta)$
$dP_y=F_ydt=\frac{kqe}{vr}sin(\theta)d\theta$
$P_y=\frac{kqe}{vr}\int_{0}^{\pi}sin(\theta)d\theta = \frac{2kqe}{vr}$
$P_y = \frac{qe}{2 \pi \epsilon_0vr}$

(b)

Let us use the answer of part (a), where instead of electron and particle q, there will be two point charges with the ratings of $$q_1,q_2$$ .

$P_y=\frac{q_1 q_2}{2 \pi \epsilon_0vr}$

(c)

$tan(\alpha) = \frac{qe}{2 \pi \epsilon_0 m v^2 r}$
$\frac{m v^2}{2}=e \Delta \phi$
$tan(\alpha) \approx \alpha$
$tan(\alpha) = \frac{kq}{\Delta \phi r} \longrightarrow r=\frac{kq}{\alpha \Delta \phi}$

Answer

(a) $P_y = \frac{qe}{2 \pi \epsilon_0vr}$

(b) $P_y=\frac{q_1 q_2}{2 \pi \epsilon_0vr}$