Question

12．如圖所示，靜止的電子在加速電壓U₁的作用下從O經(jīng)P板的小孔以速度v₁射出，運(yùn)動(dòng)時(shí)間為t₁，又垂直進(jìn)入平行金屬板間的電場(chǎng)，在偏轉(zhuǎn)電壓U₂的作用下偏轉(zhuǎn)一段距離y后離開電場(chǎng)，離開電場(chǎng)時(shí)速度為v₂，運(yùn)動(dòng)時(shí)間為t₂，則下列判斷正確的是（　　）

• A． 若只增大U₁，則t₁增大
• B． 若只增大U₁，則y減小
• C． 若只增大U₂，則v₂增大
• D． 若只增大U₂，則y減小

Answer 1

分析 電子先經(jīng)過加速電場(chǎng)加速，后經(jīng)偏轉(zhuǎn)電場(chǎng)偏轉(zhuǎn)，根據(jù)y=$\frac{{U}_{2}^{\;}{L}_{\;}^{2}}{4mlgknhb_{2}^{\;}{U}_{1}^{\;}}$可以分析電壓的變化使電子的運(yùn)動(dòng)軌跡發(fā)生相應(yīng)的變化．

解答 解：A、根據(jù)動(dòng)能定理$q{U}_{1}^{\;}=\frac{1}{2}m{v}_{1}^{2}$，得${v}_{1}^{\;}=\sqrt{\frac{2q{U}_{1}^{\;}}{m}}$，若只增大${U}_{1}^{\;}$，則${v}_{1}^{\;}$增大，由${t}_{1}^{\;}=\fracez4dxr5{\overline{v}}=\frackptoadg{\frac{{v}_{1}^{\;}}{2}}$知，${t}_{1}^{\;}$減小，故A錯(cuò)誤；
B、電子進(jìn)入偏轉(zhuǎn)電場(chǎng)${U}_{2}^{\;}$后做類平拋運(yùn)動(dòng)，$y=\frac{1}{2}a{t}_{\;}^{2}=\frac{1}{2}×\frac{q{U}_{2}^{\;}}{mvs955qk_{2}^{\;}}×（\frac{L}{{v}_{1}^{\;}}）_{\;}^{2}$=$\frac{q{U}_{2}^{\;}{L}_{\;}^{2}}{4hvlwztf_{2}^{\;}q{U}_{1}^{\;}}=\frac{{U}_{2}^{\;}{L}_{\;}^{2}}{4c1icfzl_{2}^{\;}{U}_{1}^{\;}}$，若只增大${U}_{1}^{\;}$，則y減��；若只增大${U}_{2}^{\;}$，則y增大，故B正確，D錯(cuò)誤；
C、離開電場(chǎng)時(shí)，豎直方向的分速度${v}_{y}^{\;}=at=\frac{q{U}_{2}^{\;}}{mbnrc0ux_{2}^{\;}}×\frac{L}{{v}_{1}^{\;}}=\frac{q{U}_{2}^{\;}L}{my65tw0i_{2}^{\;}}\sqrt{\frac{m}{2q{U}_{1}^{\;}}}$=$\frac{{U}_{2}^{\;}L}{5twauxa_{2}^{\;}}\sqrt{\frac{q}{2m{U}_{1}^{\;}}}$
離開時(shí)速度${v}_{2}^{\;}=\sqrt{{v}_{1}^{2}+{v}_{y}^{2}}=\sqrt{\frac{2q{U}_{1}^{\;}}{m}+\frac{{U}_{2}^{2}{L}_{\;}^{2}}{x9uk0hb_{2}^{2}}\frac{q}{2m{U}_{1}^{\;}}}$，若只增大${U}_{2}^{\;}$，則${v}_{2}^{\;}$增大，故C正確；
故選：BC

點(diǎn)評(píng) 本題考查了帶電粒子在電場(chǎng)中的運(yùn)動(dòng)，可以根據(jù)動(dòng)能定理和牛頓第二定律、運(yùn)動(dòng)學(xué)公式結(jié)合推導(dǎo)出側(cè)移y和末速度${v}_{2}^{\;}$表達(dá)式．