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

ar4senN правка от 11 июн. 2025 г., 19:17

правка #14694 предыдущая #14693 ← раньше

en/7.1.28.md +33 −3

		@@ -1,11 +1,41 @@
			### Statement

		−	~~$7.1.28.$ [Insert the problem statement]~~
		+	Electrons fly through the gap to which the potential difference $V$ is applied.
		+	Capacity of the slit length unit $C$. For small values of $V$, the angle of electron deflection by the slit field is proportional to the product of $C$ and $V$ and
		+	inversely proportional to $V_0$ ($e V_0$ is the initial electron energy): $α ≈
		+	kCV/V_0$. Determine the coefficient k.

		+	![For problem $7.1.28$\|573x262, 50%](../../img/7.1.28/photo_5355022648821807439_x.jpg)
		+
			### Solution

		−	Type your LaTeX here...
		+	From the grounding conditions (zero potential), the electron acquires velocity $v_0$, and the energy of the electron does not change.

		+	$$e V_0 = \frac{mv_0^2}{2} $$
		+
		+	The charge $q$ on the horizontal plates is derived from the capacitor equation, considering that the slit has the shape of a circle.From the grounding conditions (zero potential), the electron acquires velocity $v_0$, while the electron energy does not change.
		+
		+	$$q = 2 \pi R C V$$
		+
		+	Since the charge distribution has axial symmetry, we consider the interaction of the electron and the total charge of the horizontal plates as the interaction of two point charges.
		+
		+	Since the angle $\alpha$ is small enough under the problem conditions, we assume that the horizontal velocity changes insignificantly.We derive the charge $q$ on the horizontal plates from the capacitor equation, considering that the slit has the shape of a circle.From the grounding conditions (zero potential), the electron acquires velocity $v_0$, and the electron energy does not change.
		+
		+	$$E(\theta) = \frac{kq}{(r/sin(\theta))^2} $$\
		+	$$F(\theta)=eE(\theta)=\frac{kq}{(r/sin(\theta))^2}$$\
		+	$$dr=vdt$$\
		+	$$F_y=Fsin(\theta)$$\
		+	$$dP_y=F_ydt=\frac{kqe}{vr}sin(\theta)d\theta$$\
		+	$$P_y=\frac{kqe}{vr}\int_{0}^{\pi}sin(\theta)d\theta = \frac{2kqe}{vr}$$\
		+	$$P_x = mv$$\
		+	$$tan(\alpha)=\frac{P_y}{P_x} = \frac{qe}{2 \pi \epsilon_0 m v_0^2 R}$$\
		+	$$\alpha \approx \frac{qe}{2 \pi \epsilon_0 m v_0^2 R}$$
		+
		+	Now substitute the value of $q,e$ and compare with the formula from the problem.
		+
		+	$$\alpha \approx \frac{CV}{2 \epsilon_0 V_0}$$\
		+	$$k=\frac{1}{2 \epsilon_0}$$
		+
			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$$k=\frac{1}{2 \epsilon_0}$$

	@@ -1,11 +1,41 @@
	### Statement		### Statement

	~~$7.1.28.$ [Insert the problem statement]~~		Electrons fly through the gap to which the potential difference $V$ is applied.
			Capacity of the slit length unit $C$. For small values of $V$, the angle of electron deflection by the slit field is proportional to the product of $C$ and $V$ and
			inversely proportional to $V_0$ ($e V_0$ is the initial electron energy): $α ≈
			kCV/V_0$. Determine the coefficient k.

			![For problem $7.1.28$\|573x262, 50%](../../img/7.1.28/photo_5355022648821807439_x.jpg)

	### Solution		### Solution

	Type your LaTeX here...		From the grounding conditions (zero potential), the electron acquires velocity $v_0$, and the energy of the electron does not change.

			$$e V_0 = \frac{mv_0^2}{2} $$

			The charge $q$ on the horizontal plates is derived from the capacitor equation, considering that the slit has the shape of a circle.From the grounding conditions (zero potential), the electron acquires velocity $v_0$, while the electron energy does not change.

			$$q = 2 \pi R C V$$

			Since the charge distribution has axial symmetry, we consider the interaction of the electron and the total charge of the horizontal plates as the interaction of two point charges.

			Since the angle $\alpha$ is small enough under the problem conditions, we assume that the horizontal velocity changes insignificantly.We derive the charge $q$ on the horizontal plates from the capacitor equation, considering that the slit has the shape of a circle.From the grounding conditions (zero potential), the electron acquires velocity $v_0$, and the electron energy does not change.

			$$E(\theta) = \frac{kq}{(r/sin(\theta))^2} $$\
			$$F(\theta)=eE(\theta)=\frac{kq}{(r/sin(\theta))^2}$$\
			$$dr=vdt$$\
			$$F_y=Fsin(\theta)$$\
			$$dP_y=F_ydt=\frac{kqe}{vr}sin(\theta)d\theta$$\
			$$P_y=\frac{kqe}{vr}\int_{0}^{\pi}sin(\theta)d\theta = \frac{2kqe}{vr}$$\
			$$P_x = mv$$\
			$$tan(\alpha)=\frac{P_y}{P_x} = \frac{qe}{2 \pi \epsilon_0 m v_0^2 R}$$\
			$$\alpha \approx \frac{qe}{2 \pi \epsilon_0 m v_0^2 R}$$

			Now substitute the value of $q,e$ and compare with the formula from the problem.

			$$\alpha \approx \frac{CV}{2 \epsilon_0 V_0}$$\
			$$k=\frac{1}{2 \epsilon_0}$$

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$$k=\frac{1}{2 \epsilon_0}$$