Question

圖中y軸上A、B兩點(diǎn)的縱坐標(biāo)分別為d和-d.在0≤y≤d的區(qū)域中,存在沿y軸向上的非均勻電場(chǎng),場(chǎng)強(qiáng)E的大小與y成正比,即E=ky;在y≥d的區(qū)域中,存在沿 y軸向上的勻強(qiáng)電場(chǎng),電場(chǎng)強(qiáng)度E=kd(k屬未知量).x軸下方空間各點(diǎn)電場(chǎng)分布與x軸上方空間中的分布對(duì)稱,只是場(chǎng)強(qiáng)的方向都沿著y軸向下.現(xiàn)有一帶電荷量為q、質(zhì)量為m的微粒甲正好在O、B兩點(diǎn)之間做簡(jiǎn)諧運(yùn)動(dòng).某時(shí)刻將一帶電荷量為2q、質(zhì)量為m的微粒乙從y軸上的C點(diǎn)處由靜止釋放,乙運(yùn)動(dòng)到O點(diǎn)和甲相碰并結(jié)為一體(忽略兩微粒之間的庫侖力).在以后的運(yùn)動(dòng)中,它們所能達(dá)到的最高點(diǎn)和最低點(diǎn)分別為A點(diǎn)和D點(diǎn),且經(jīng)過P點(diǎn)時(shí)速度達(dá)到最大值(重力加速度為g).

(1)求勻強(qiáng)電場(chǎng)E;

(2)求出AB間的電勢(shì)差U_AB及OB間的電勢(shì)差U_OB;

(3)分別求出P、C、D三點(diǎn)到O點(diǎn)的距離.

Answer 1

(1)甲做簡(jiǎn)諧運(yùn)動(dòng)到達(dá)最高點(diǎn)時(shí)和到達(dá)最低點(diǎn)時(shí)加速度相等,所受合力大小也相等,則:

qE_勻-mg=mg                                                          

    E_勻=2mg/q.                                                           

(2)U_AB=0                                                              

    mgd-qU_OB=0                                                           

    U_OB=mgd/q.                                                         

(3)它們經(jīng)過P點(diǎn)時(shí)速度最大，所受合力為0

    3qE_P=2mg                                                           

    E_P=2mg/3q=E_勻/3即kh_OP=kd/3  h_OP=d/3                                   

    微粒甲在從C到O的過程中：2qE_勻（h_CO-d）+2qU_OA+mgh_CO=E_kO甲-0       

    微粒甲和微粒乙在碰撞過程中近似動(dòng)量守恒,則有:E_kO甲∶E_kA=(p²/2m)∶(p²/4m)

    結(jié)合體在從O到A過程中:-2mgd-3qU_OA=0-E_kO                           

    結(jié)合體在從O到D過程中:-3qE_勻(h_OD-d)-3qU_OB+2mgh_OD=0-E_kO              

    解得:h_CO=2.4d                                                        

    h_OD=2d.