ÌâÄ¿ÄÚÈÝ

11£®ÈçͼËùʾ£¬Ë®Æ½·ÅÖõÄÁ½¿éƽÐнðÊô°å£¬Ïà¾àd=0.04m£¬Á½°å¼äµÄµçѹÊÇU=1.2¡Á103V£¬ÔÚÁ½°å¼äµÄµç³¡ÖÐÓÐÒ»ÖÊÁ¿ÊÇm=6.0¡Á10-3kgµÄ´øµçСÇò£¬¾²Ö¹ÓÚÆ½ÐаåµÄÖе㣬ȡg=10m/s2£¬ÊÔÇó£º
£¨1£©Ð¡Çò´øºÎÖÖµçºÉ£»
£¨2£©Á½¿éƽÐнðÊô°åÖ®¼äµÄ³¡Ç¿£»
£¨3£©Ð¡ÇòËù´øµçÁ¿£»
£¨4£©Á½°å¼äµÄµçѹ±£³Ö²»±ä£¬°ÑÁ½°å¼äµÄ¾àÀë¸Ä±äΪ0.08m£¬Ð¡Çò½«ÏòÄĸö·½ÏòÔ˶¯£¿¼ÓËÙ¶ÈÊǶàÉÙ£¿

·ÖÎö ´øµçÒºµÎ´¦ÓÚ¾²Ö¹×´Ì¬£¬ÊÜÁ¦Æ½ºâ£¬ÖØÁ¦Óëµç³¡Á¦´óСÏàµÈ£¬·½ÏòÏà·´£¬¸ù¾ÝƽºâÌõ¼þÁÐʽ£¬Çó½â°å¼äµÄµç³¡Ç¿¶È£¬Óй²µãÁ¦Æ½ºâÇóµÄµçºÉÁ¿£¬ÓÉÅ£¶ÙµÚ¶þ¶¨ÂÉÇóµÄ¼ÓËÙ¶È

½â´ð ½â£º£¨1£©¸ù¾ÝÁ¦µÄƽºâÌõ¼þ£¬ÓÉmg=qE£¬Êܵ½µÄµç³¡Á¦ÏòÉÏ£¬µç³¡ÏòÏ£¬¹ÊСÇòµã¸ºµç
£¨2£©µç³¡Ç¿¶ÈΪ£º$E=\frac{U}{d}=\frac{1200}{0.04}V/m=3¡Á1{0}^{4}V/m$
£¨3£©Óй²µãÁ¦Æ½ºâ¿ÉÖª£ºmg=qE
q=$\frac{mg}{E}=\frac{6¡Á1{0}^{-3}¡Á10}{3¡Á1{0}^{4}}C=2¡Á1{0}^{-6}C$
£¨4£©ÓÉE=$\frac{U}{d}$¿ÉÖª£¬µ±dÔö´ó£¬µç³¡Ç¿¶È¼õС£¬Êܵ½µÄµç³¡Á¦¼õССÓÚÖØÁ¦ÏòϼÓËÙÔ˶¯£¬ÓÉÅ£¶ÙµÚ¶þ¶¨Âɵãº
$mg-\frac{qU}{d¡ä}=ma$
´úÈëÊý¾Ý½âµÃ£ºa=5m/s2
´ð£º£¨1£©Ð¡Çò´ø¸ºµç£»
£¨2£©Á½¿éƽÐнðÊô°åÖ®¼äµÄ³¡Ç¿3¡Á104V/m£»
£¨3£©Ð¡ÇòËù´øµçÁ¿2¡Á10-6C£»
£¨4£©Á½°å¼äµÄµçѹ±£³Ö²»±ä£¬°ÑÁ½°å¼äµÄ¾àÀë¸Ä±äΪ0.08m£¬Ð¡Çò½«ÏòÏÂÔ˶¯£¬¼ÓËÙ¶ÈÊÇ5m/s2

µãÆÀ ±¾ÌâÊÇ´øµçÌåÔڵ糡ÖÐÆ½ºâÎÊÌâ£¬ÕÆÎÕÆ½ºâÌõ¼þºÍµç³¡Ç¿¶ÈÓëÊÔ̽µçºÉÎ޹صÄÌØµã½øÐÐÇó½â£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø