Question

14£®ÎïÀíѧÖÐÓÐЩÎÊÌâµÄ½áÂÛ²»Ò»¶¨±ØÐëͨ¹ý¼ÆËã²ÅÄÜÑéÖ¤£¬ÓÐʱֻÐèҪͨ¹ýÒ»¶¨µÄ·ÖÎö¾Í¿ÉÒÔÅжϽáÂÛÊÇ·ñÕýÈ·£®ÈçͼËùʾΪÁ½¸ö±Ë´ËƽÐÐÇÒ¹²ÖáµÄ°ë¾¶·Ö±ðΪR₁ºÍR₂µÄÔ²»·£¬Á½Ô²»·ÉϵĵçºÉÁ¿¾ùΪq£¨q£¾0£©£¬¶øÇÒµçºÉ¾ùÔÈ·Ö²¼£®Á½Ô²»·µÄÔ²ÐÄO₁ºÍO₂Ïà¾àΪ2a£¬ÁªÏßµÄÖеãΪO£¬ÖáÏßÉϵÄAµãÔÚOµãÓÒ²àÓëOµãÏà¾àΪr£¨r£¼a£©£®ÊÇ·ÖÎöÅжÏÏÂÁйØÓÚAµã´¦µç³¡Ç¿¶È´óСEµÄ±í´ïʽ£¨Ê½ÖÐkΪ¾²µçÁ¦³£Á¿£©ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

• A£® E=|$\frac{kq{R}_{1}}{[{{R}_{1}}^{2}+£¨a+r£©^{2}]}$-$\frac{kq{R}_{2}}{[{{R}_{2}}^{2}+£¨a-r£©^{2}]}$|
• B£® E=|$\frac{kq{R}_{1}}{[{{R}_{1}}^{2}+£¨a+r£©^{2}]^{\frac{3}{2}}}$-$\frac{kq{R}_{2}}{[{{R}_{2}}^{2}+£¨a-r£©^{2}]^{\frac{3}{2}}}$|
• C£® E=|$\frac{kq£¨a+r£©}{[{{R}_{1}}^{2}+£¨a+r£©^{2}]}$-$\frac{kq£¨a-r£©}{{{[R}_{2}}^{2}+£¨a-r£©^{2}]}$|
• D£® E=|$\frac{kq£¨a+r£©}{[{{R}_{1}}^{2}+£¨a+r£©^{2}]^{\frac{3}{2}}}$-$\frac{kq£¨a-r£©}{[{{R}_{2}}^{2}+£¨a-r£©^{2}]^{\frac{3}{2}}}$|

Answer 1

·ÖÎö ÌâÄ¿ÒªÇó²»Í¨¹ý¼ÆË㣬ֻÐèͨ¹ýÒ»¶¨µÄ·ÖÎö¾Í¿ÉÒÔÅжϽáÂÛ£¬ËùÒÔ¸ù¾ÝµãµçºÉ³¡Ç¿µÄ¹«Ê½E=k$\frac{Q}{{r}^{2}}$£¬ÓëÑ¡ÏîÏà¶Ô±È£¬Ñ°ÕÒ²»Í¬µã£¬ÔÙÓü«ÏÞ·ÖÎöÎÊÌâµÄ˼Ïë·½·¨¾Í¿ÉÒÔ·ÖÎö³ö½á¹û£®

½â´ð ½â£ºÓëµãµçºÉµÄ³¡Ç¿¹«Ê½E=k$\frac{Q}{{r}^{2}}$£¬±È½Ï¿ÉÖª£¬A¡¢CÁ½Ïî±í´ïʽµÄµ¥Î»²»Êdz¡Ç¿µÄµ¥Î»£¬¹Ê¿ÉÒÔÅųýAC£»
µ±r=aʱ£¬ÓÒ²àÔ²»·ÔÚAµã²úÉúµÄ³¡Ç¿ÎªÁ㣬ÔòA´¦³¡Ç¿Ö»ÓÉ×ó²àÔ²»·ÉϵĵçºÉ²úÉú£¬¼´³¡Ç¿±í´ïʽֻÓÐÒ»Ï¹ÊB´íÎó£®×ÛÉÏËùÊö£¬¿ÉÖªDÕýÈ·£®
¹ÊÑ¡£ºD£®

µãÆÀ ±¾Ì⿼²é¶¨ÐÔ¡¢¼«ÏÞ·ÖÎöÎÊÌâµÄ˼Ïë·½·¨£¬ÒªÇóѧÉúÔÚÀι̵ÄÕÆÎÕ»ù±¾ÖªÊ¶µÄ»ù´¡ÉÏÒªÄܹýÁé»îµÄ·ÖÎöÎÊÌ⣮