ÌâÄ¿ÄÚÈÝ

ÈçͼËùʾΪÃÜÁ¢¸ùÓ͵ÎʵÑéʾÒâͼ¡£ÔÚµç½éÖÊΪ¿ÕÆøµÄµçÈÝÆ÷ÖУ¬¹Û²âÒÔijһËٶȵÎÈëµÄÓ͵Ρ£µ±¿ª¹ØS¶Ï¿ªÊ±£¬´ÓÉÏ°åС¿×Æ®ÈëµÄ´øµçÓ͵ÎÄÜÒÔÎȶ¨µÄËÙÂÊv1Ͻµ¡£ºÏÉÏS£¬¹ýÒ»»á¶ùÓ͵ÎÓÉϽµ×ªÎªÉÏÉý£¬Îȶ¨Ê±ËÙÂÊΪv2¡£ÉèÓ͵εÄÖÊÁ¿Îªm£¬µçÔ´µçѹΪU£¬°å¼ä¾àÀëΪd£¬Ó͵ÎÊܵ½¿ÕÆøµÄð¤ÖÍ×èÁ¦µÄ´óСÓëËٶȳÉÕý±È£¬¼´Ff=kv(ʽÖÐkΪδ֪Á¿)¡£

(1)ÇóÓ͵ÎËù´øµÄµçºÉÁ¿q¡£

(2)ÉèÁ½°å¼ä¾àd=0.5 cm£¬°å¼äµçѹU=150 V£¬²âµÃÓ͵εÄÖ±¾¶D=1.10¡Á10-6 m£¬Ó͵εÄÃܶȦÑ=1.05¡Á103 kg/m3£¬ÈôʵÑéÖй۲쵽Ó͵ÎÏòϺÍÏòÉÏÔÈËÙÔ˶¯µÄËÙÂÊÏàµÈ£¬ÊÔÓɴ˼ÆËãÓ͵εĵçºÉÁ¿²¢ËµÃ÷µçÐÔ¡£

¡¾½âÌâÖ¸ÄÏ¡¿½â´ð±¾ÌâʱӦעÒâÒÔÏÂÈýµã£º

(1)ÕýÈ··ÖÎöÓ͵εÄÊÜÁ¦Çé¿öºÍÔ˶¯×´Ì¬£»

(2)×èÁ¦´óСFf=kvÖеÄkÊÇδ֪Êý£¬´ð°¸Öв»ÄÜÓÐk£»

(3)ÕýÈ·±í´ï³öÓ͵ÎÖÊÁ¿ÓëÖ±¾¶µÄ¹Øϵʽ¡£

¡¾½âÎö¡¿(1)ÉèÏòÏÂΪÕý·½Ïò£¬SºÏÉÏÇ°£¬Ó͵ÎÒÔËÙÂÊv1ÔÈËÙϽµ£¬ÊÜÖØÁ¦mgºÍÏòÉϵĿÕÆøð¤ÖÍ×èÁ¦kv1×÷Óã¬ÓÐ:mg=kv1                                 (3·Ö)

SºÏÉϺóÓ͵ÎÒÔËÙÂÊv2ÔÈËÙÉÏÉýʱ£¬µç³¡Á¦qE±ØÏòÉÏ£¬ÖØÁ¦mgºÍ¿ÕÆø×èÁ¦kv2¾ùÏòÏ£¬ÓУºqE=mg+kv2

ÓÖE£½                                                                            (3·Ö)

ÓÉÉÏÊöÈýʽ½âµÃ£º¡£                                         (2·Ö)

(2)ÉèÓ͵εÄËÙÂÊΪv0£¬Ï½µÊ±ÓÐmg=kv0£¬                                   (1·Ö)

ÉÏÉýʱÓÐqE=mg+kv0£¬                                                         (1·Ö)

ÓÖE=£¬                                                                           (1·Ö)

Ó͵εÄÖÊÁ¿m=¦ÑV=£¬½âµÃq=£¬                                 (2·Ö)

´úÈëÊý¾Ý¿É½âµÃq¡Ö4.9¡Á10-19 C¡£

Ó͵ÎÓ¦´ø¸ºµç¡£                                                                  (2·Ö)

´ð°¸:(1)

(2)4.9¡Á10-19 C      Ó͵δø¸ºµç

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø