A point charge of 2.4 $2\mathrm{.}4$mu C is 0.50$0\mathrm{.}50$ cm away from a second point charge of +2.0 $+2\mathrm{.}0$mu C.$.$ The force on a third change, +1.4 $+1\mathrm{.}4$mu C,$,$ placed exactly halfway between the other two, is:

 ($($a)0.88$)0\mathrm{.}88$ N,$,$ directed toward the negative charge;

 ($($b)0.22$)0\mathrm{.}22$ N,$,$ directed toward the negative charge

 ($($c)3.32$)3\mathrm{.}32$ N,$,$ directed toward the negative charge

 ($($d)0.22$)0\mathrm{.}22$ N,$,$ directed toward the positive change. 

 

Answer rating: 100% (QA)

Solution finding the distance between the third and the se... View the full answer