Правка разделов «Statement», «Solution», «Answer»

ar4senN правка от
Посмотреть решение История
правка #14664 предыдущая #14663 ← раньше
en/7.1.27.md +53 −3
@@ -1,11 +1,61 @@
### Statement
−$7.1.27.$ [Insert the problem 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.
+
+![For problem $7.1.27$|985x251, 50%](../../img/7.1.27/photo_5346012945601132071_y.jpg)
+
+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
−$E$
+(a)
+Since the electric field is axisymmetric, we will use superposition, where an electron interacts with a point charge q.
+
+![|948x550, 50%](../../img/7.1.27/photo_5346012945601132056_y.jpg).
+
+$$E(\theta) = \frac{kq}{(r/sin(\theta))^2} $$\
+$$F(\theta)=eE(\theta)=\frac{kq}{(r/sin(\theta))^2}$$
+$$dr=vdt$$\
+$$F_y=Fsin(\theta)$$\
+$$dP_y=F_ydt=\frac{kqe}{vr}sin(\theta)d\theta$$\
+$$P_y=\frac{kqe}{vr}\int_{0}^{\pi}sin(\theta)d\theta = \frac{2kqe}{vr}$$\
+$$P_y = \frac{qe}{2 \pi \epsilon_0vr}$$
+
+(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
−[Insert a concise answer or boxed result]
+(a) $$P_y = \frac{qe}{2 \pi \epsilon_0vr}$$
+
+(b) $$P_y=\frac{q_1 q_2}{2 \pi \epsilon_0vr}$$
+
+(c) $$r=\frac{kq}{\alpha \Delta \phi}=1 \cdot 10^{\text -10} (m)$$