Question #126622

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.278 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.62 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, the answer to part (a) being the one with the greater (and positive) value?


1
Expert's answer
2020-07-20T14:57:05-0400

Initially, WLOG let q1q2q_1\geq q_2 ,thus magnitude is

F=kq1q2r2    q1q2=r2FkF=k\frac{q_1q_2}{r^2}\implies q_1q_2=\frac{r^2F}{k}

Suppose qq charge is shared ,thus q=q1q22q=\frac{q_1-q_2}{2} ,hence charge on each object are q1+q22\frac{q_1+q_2}{2} ,Therefore

F=k(q1+q22)2r2    q1+q2=±4r2Fk=±2q1q2    (q1q2)2=0    q1=q2F=k\frac{(\frac{q_1+q_2}{2})^2}{r^2}\implies q_1+q_2=\pm\sqrt{\dfrac{4r^2F}{k}}=\pm2\sqrt{q_1q_2}\\ \implies (\sqrt{q_1}\mp\sqrt{q_2})^2=0\implies q_1=q_2

Hence,

q1=r2Fk=2.31μCq2=2.31μCq_1=\sqrt{\frac{r^2F}{k}}=2.31\mu C\\q_2=-2.31\mu C


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS