ÌâÄ¿ÄÚÈÝ

9£®ÈçͼËùʾ£¬Õæ¿ÕÖй̶¨ÔÚOµãµÄµãµçºÉ´øµçÁ¿Q=+2.0¡Á10-6C£¬ÐéÏßΪÁíÒ»´øµçÁ¿q=-2.0¡Á10-9CµÄµãµçºÉ´ÓÎÞÇîÔ¶´¦ÏòOÔ˶¯µÄ¹ì¼££®µãµçºÉq´ÓÎÞÇîÔ¶´¦ÒƵ½Aµã¾²µçÁ¦×öÁË1.5¡Á10-7JµÄ¹¦£»È¡ÎÞÇîÔ¶´¦µçÊÆÎª0£¬¹ì¼£ÉÏÀëOµã¾àÀëΪ3cmµÄBµãµçÊÆ¦ÕB=125V£¬¾²µçÁ¦³£Á¿k=9.0¡Á109N•m2/C2£®Çó£º
£¨1£©µãµçºÉqÔÚBµãʱÊܵ½µÄ¾²µçÁ¦F´óС£»
£¨2£©µãµçºÉqÔÚAµãµÄµçÊÆÄÜEpA£»
£¨3£©A¡¢BÁ½µã¼äµÄµçÊÆ²îUAB£®

·ÖÎö £¨1£©¸ù¾Ý¿âÂØ¶¨ÂɼÆË㣻
£¨2£©¸ù¾Ý¹¦ÄܹØÏµ£¬¾²µçÁ¦×öµÄ¹¦µÈÓÚµçÊÆÄܱ仯Á¿Çó½â£»
£¨3£©¸ù¾ÝµçÊÆ²îµÄ¶¨Òåʽ¼ÆË㣮

½â´ð ½â£º
£¨1£©µãµçºÉqÔÚBµãʱÊܵ½µÄ¾²µçÁ¦´óС$F=k\frac{Qq}{r^2}$
½âµÃF=4¡Á10-2N
 £¨2£©¸ù¾Ý¹¦ÄܹØÏµ£ºW¡ÞA=EP¡Þ-EPA
½âµÃ ${E_{PA}}=-1.5¡Á{10^{-7}}J$
£¨3£©AµãµÄµçÊÆ    ${¦Õ_A}=\frac{{{E_{PA}}}}{q}=\frac{{-1.5¡Á{{10}^{-7}}}}{{-2¡Á{{10}^{-9}}}}=75V$
A¡¢BÁ½µãµÄµçÊÆ²îUABAB
½âµÃ UAB=-50V
´ð£º£¨1£©µãµçºÉqÔÚBµãʱÊܵ½µÄ¾²µçÁ¦´óСΪF=4¡Á10-2N£»
£¨2£©µãµçºÉqÔÚAµãµÄµçÊÆÄÜΪ ${E_{PA}}=-1.5¡Á{10^{-7}}J$£»
£¨3£©A¡¢BÁ½µã¼äµÄµçÊÆ²îΪ50V£®

µãÆÀ ½â´ËÌâµÄ¹Ø¼üÊÇÒªÊìϤ¿âÂØ¶¨ÂÉ£¬µçÊÆÄÜ£¬µçÊÆ²î¸ÅÄÒ×´íµãÔÚ£¨2£©ÎÊÖÐÀûÓþ²µçÁ¦×ö¹¦ÓëµçÊÆÄܱ仯µÄ¹ØÏµ¼ÆËãµçÊÆÄÜ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø