ÌâÄ¿ÄÚÈÝ

18£®Èçͼ£¬¾øÔµµÄË®Æ½ÃæÉÏ£¬Ïà¸ô2LµÄABÁ½µã¹Ì¶¨ÓÐÁ½¸öµçÁ¿¾ùΪQµÄÕýµãµçºÉ£¬a£¬O£¬bÊÇABÁ¬ÏßÉϵÄÈýµã£¬ÇÒOΪÖе㣬Oa=Ob=$\frac{L}{2}$£®Ò»ÖÊÁ¿Îªm¡¢µçÁ¿Îª+qµÄµãµçºÉÒÔ³õËÙ¶Èv0´Óaµã³ö·¢ÑØABÁ¬ÏßÏòBÔ˶¯£¬ÔÚÔ˶¯¹ý³ÌÖеçºÉÊܵ½´óСºã¶¨µÄ×èÁ¦×÷Ó㬵±ËüµÚÒ»´ÎÔ˶¯µ½OµãʱËÙ¶ÈΪ2v0£¬¼ÌÐøÔ˶¯µ½bµãʱËٶȸպÃΪÁ㣬Ȼºó·µ»Ø£¬×îºóǡͣÔÚOµã£®ÒÑÖª¾²µçÁ¦ºãÁ¿Îªk£®Çó£º
£¨1£©aµãµÄ³¡Ç¿´óС£®
£¨2£©×èÁ¦µÄ´óС£®
£¨3£©aOÁ½µã¼äµÄµçÊÆ²î£®
£¨4£©µçºÉÔڵ糡ÖÐÔ˶¯µÄ×Ü·³Ì£®

·ÖÎö £¨1£©Ó¦ÓõãµçºÉµÄ³¡Ç¿¹«Ê½Ó볡µÄµþ¼ÓÔ­ÀíÇó³öµç³¡Ç¿¶È£®
£¨2£©Óɶ¯Äܶ¨ÀíÇó³ö×èÁ¦£®
£¨3£©Ó¦Óö¯Äܶ¨ÀíÇó³öµçÊÆ²î£¬È»ºó¸ù¾ÝµçÊÆ²îµÄ¶¨ÒåʽÇó³öµçÊÆ£®
£¨4£©Ó¦Óö¯Äܶ¨ÀíÇó³ö·³Ì£®

½â´ð ½â£º
£¨1£©ÓɵãµçºÉµç³¡Ç¿¶È¹«Ê½ºÍµç³¡µþ¼ÓÔ­Àí¿ÉµÃ£º
${E}_{a}=\frac{kQ}{£¨\frac{L}{2}£©^{2}}-\frac{kQ}{£¨\frac{3L}{2}£©^{2}}=\frac{32kQ}{9{L}^{2}}$
£¨2£©´Óaµ½bµã¹ý³ÌÖУ¬¸ù¾Ý¶Ô³ÆÐÔ£¬Ua=Ub¸ù¾Ý¶¯Äܶ¨Àí$-fL=0-\frac{1}{2}m{v_0}^2$
$f=\frac{1}{2L}m{v_0}^2$
£¨3£©´Óaµ½oµã¹ý³ÌÖУ¬¸ù¾Ý¶¯Äܶ¨Àí$q{U_{ao}}-f\frac{L}{2}=\frac{1}{2}m{£¨2{v_0}£©^2}-\frac{1}{2}m{v_0}^2$
${U_{ao}}=\frac{7mv_0^2}{4q}$
£¨4£©×îºóÍ£ÔÚoµã£¬Õû¸ö¹ý³Ì·ûºÏ¶¯Äܶ¨Àí$q{U_{ao}}-fS=0-\frac{1}{2}m{v_0}^2$
$S=\frac{9}{2}L$=4.5L
´ð£º£¨1£©aµãµÄµç³¡Ç¿¶È$\frac{32kQ}{9{L}^{2}}$£®
£¨2£©µçºÉqÊܵ½×èÁ¦µÄ´óСÊÇ$\frac{1}{2L}m{{v}_{0}}^{2}$£®
£¨3£©aµãµÄµçÊÆ$\frac{7m{v}_{0}^{2}}{4q}$£®
£¨4£©µçºÉqÔڵ糡ÖÐÔ˶¯µÄ×Ü·³Ì4.5L£®

µãÆÀ ±¾Ì⿼²éÁ˶¯Äܶ¨ÀíµÄÓ¦Ó㬷ÖÎöÇå³þµçºÉµÄÔ˶¯¹ý³Ì£¬Ó¦Óö¯Äܶ¨Àí¡¢µãµçºÉµÄ³¡Ç¿¹«Ê½Ó볡µÄµþ¼ÓÔ­Àí¼´¿ÉÕýÈ·½âÌ⣮

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø