Question

1£®ÈçͼËùʾ£¬CΪƽÐаåµçÈÝÆ÷£¬ÐéÏßµ½Á½¼«°å¾àÀëÏàµÈ£¬¼«°å³¤L₁=8.0¡Á10^-2m£¬Á½°å¼ä¾àd=1.0¡Á10^-2m£®SΪÆÁ£¬Ó뼫°å´¹Ö±£¬µ½¼«°åµÄ¾àÀëL₂=0.16m£®ÓÐһϸµç×ÓÊøÑØͼÖÐÐéÏßÒÔËÙ¶ÈV₀=8.0¡Á10⁶m/sÁ¬Ðø²»¶ÏµØÉäÈëC£®ÒÑÖªµç×ÓµçºÉÁ¿e=1.6¡Á10^-19C£¬µç×ÓÖÊÁ¿m=9.0¡Á10^-31kg£®ºöÂÔµç×ÓËùÊܵÄÖØÁ¦£¬²»¿¼ÂǵçÈÝÆ÷±ßÔµµç³¡µÄ±ä»¯£¬µçÈÝÆ÷ÍⲿµÄµç³¡ÎªÁ㣮
£¨1£©µ±µçÈÝÆ÷µÄµçÊƲîU₁=5.4V£¬Åжϵç×ÓÊÇ·ñÄܹ»´©¹ýµçÈÝÆ÷£®
£¨2£©µ±µçÈÝÆ÷µÄµçÊƲîU₂=2.7V£¬Çóµç×Óµ½´ïÆÁSÉÏʱ£¬ËüÀëOµãµÄ¾àÀ룮

Answer 1

·ÖÎö £¨1£©µ±µçÈÝÆ÷µÄµçÊƲîU₁=5.4Vʱ£¬µç×Ó½øÈëµç³¡ÖÐ×öÀàƽÅ×Ô˶¯£¬¼ÙÉèÄܹ»Éä³ö£¬ÔËÓÃÔ˶¯µÄÕý½»·Ö½â·¨£¬¸ù¾Ý·ÖÔ˶¯¹«Ê½ÁÐʽÇó½â²àÒÆÁ¿½øÐÐÅжϼ´¿É£»
£¨2£©µ±µçÈÝÆ÷µÄµçÊƲîU₂=2.7V£¬µç×Ó½øÈëµç³¡ÖÐ×öÀàƽÅ×Ô˶¯£¬À뿪µç³¡ºó×öÔÈËÙÖ±ÏßÔ˶¯£»¶ÔÀàƽÅ×Ô˶¯¹ý³Ì£¬¸ù¾Ý·ÖÔ˶¯¹«Ê½Çó½âËÙ¶Èƫת½ÇµÄÕýÇÐÖµ£¬»­³ö¹ì¼££¬½áºÏ¼¸ºÎ¹Øϵ½øÐзÖÎöÇó½â£®

½â´ð ½â£º£¨1£©µçÈÝÆ÷µÄµçÊƲîU₁=5.4Vʱ£¬¼ÙÉèµç×ÓÄܹ»Éä³ö£¬¸ù¾ÝÀàËÆƽÅ×Ô˶¯µÄ·ÖÔ˶¯¹«Ê½£¬ÓУº
L₁=v₀t
$y=\frac{1}{2}a{t}^{2}$ 
ÆäÖУº
a=$\frac{e{U}_{1}}{md}$
ÁªÁ¢½âµÃ£º
$y=\frac{{e{U_1}L_1^2}}{2mdv_0^2}=\frac{{1.6¡Á1{0^{-19}}¡Á5.4¡Á{{£¨8¡Á1{0^{-2}}£©}^2}}}{{2¡Á9¡Á1{0^{-31}}¡Á1¡Á1{0^{-2}}¡Á{{£¨8¡Á1{0^6}£©}^2}}}=4.8¡Á1{0^{-3}}m$
ÓÉÓÚy$£¼\frac{d}{2}$£¬¹ÊÄܹ»Éä³ö¼«°å£»
£¨2£©µçÈÝÆ÷µÄµçÊƲîU₂=2.7V£¬¶ÔÀàƽÅ×Ô˶¯¹ý³Ì£¬ÓУº
v_x=v₀
v_y=at 
ÆäÖУº
a=$\frac{e{U}_{2}}{md}$
¹ÊËÙ¶Èƫת½ÇµÄÕýÇÐֵΪ£º
tan¦È=$\frac{{v}_{y}}{{v}_{x}}$=$\frac{e{U}_{2}L}{md{v}_{0}^{2}}$
»­³öÔ˶¯¹ì¼££¬ÈçͼËùʾ£º

½áºÏ¼¸ºÎ¹Øϵ£¬ÓУº
y=£¨$\frac{{L}_{1}}{2}+{L}_{2}$£©tan¦È
ÁªÁ¢½âµÃ£º
y=£¨$\frac{{L}_{1}}{2}+{L}_{2}$£©$\frac{e{U}_{2}L}{md{v}_{0}^{2}}$=$£¨\frac{1}{2}¡Á8¡Á1{0^{-2}}+0.16£©¡Á\frac{{1.6¡Á1{0^{-19}}¡Á2.7¡Á£¨8¡Á1{0^{-2}}£©}}{{9¡Á1{0^{-31}}¡Á1¡Á1{0^{-2}}¡Á{{£¨8¡Á1{0^6}£©}^2}}}=1.2¡Á1{0^{-2}}m$
´ð£º£¨1£©µ±µçÈÝÆ÷µÄµçÊƲîU₁=5.4V£¬µç×ÓÄܹ»´©¹ýµçÈÝÆ÷£®
£¨2£©µ±µçÈÝÆ÷µÄµçÊƲîU₂=2.7V£¬µç×Óµ½´ïÆÁSÉÏʱ£¬ËüÀëOµãµÄ¾àÀëΪ1.2¡Á10^-2m£®

µãÆÀ ±¾Ìâ×ÅÖØ¿¼²é´øµçÁ£×ÓÔڵ糡ÖеÄƫתÎÊÌ⣬ҪÊìÁ·ÔËÓÃÔ˶¯µÄ·Ö½â·¨Ñо¿ÀàƽÅ×Ô˶¯£¬ÄÜÓÉÅ£¶ÙµÚ¶þ¶¨ÂÉ¡¢Ô˶¯Ñ§¹«Ê½¼°Êýѧ֪ʶµÃµ½×ÛºÏʽ£¬ÔÙ½øÐÐÇó½â£¬¹ý³ÌÖ®Öпɲ»´úÊý¾Ý£¬´Ó¶ø¼õС¼ÆËãÁ¿£®