Two objects are identical and small enough that their sizes can be ignored relat

Question

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.21 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 2.7 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? (Note: there are two possible pairs of answers, but assume q1 to be the larger number.)

Explanation / Answer

Thisis a little tricky...   lets start with twocharges    Q and q   such that Q ispositive (and larger) and q is negative (and smaller). Then weknow...     - F = k Q q /d2      or     Qq = - Fd2 / k = - 2.7 * 0.212 /8.99 x 109       Qq   =   0.0132 x10-7
     Now... when they are brought together, each object gets half thetotal, or    charge on each is now    ( Q + q) /2      and then the force is    F = k [ (Q + q)/2 ]2 / d2      or     [(Q + q)/2 ]2     =  F d2 /k   =   0.0132 x10-7 Takethe square root of both sides and multiply by 2 and youget             Q + q =    2.30 x10-6 Now...if we measure the charges in microCoulomb we can eliminate theexponents and     Q+ q = 2.30 x10-7         Qq = 0.0132 x10-7 twoequations, two unknowns... solve! Thisis a little tricky...   lets start with twocharges    Q and q   such that Q ispositive (and larger) and q is negative (and smaller). Then weknow...     - F = k Q q /d2      or     Qq = - Fd2 / k = - 2.7 * 0.212 /8.99 x 109       Qq   =   0.0132 x10-7
     Now... when they are brought together, each object gets half thetotal, or    charge on each is now    ( Q + q) /2      and then the force is    F = k [ (Q + q)/2 ]2 / d2      or     [(Q + q)/2 ]2     =  F d2 /k   =   0.0132 x10-7 Takethe square root of both sides and multiply by 2 and youget             Q + q =    2.30 x10-6 Now...if we measure the charges in microCoulomb we can eliminate theexponents and     Q+ q = 2.30 x10-7         Qq = 0.0132 x10-7 twoequations, two unknowns... solve! Then weknow...     - F = k Q q /d2      or     Qq = - Fd2 / k = - 2.7 * 0.212 /8.99 x 109       Qq   =   0.0132 x10-7
     Now... when they are brought together, each object gets half thetotal, or    charge on each is now    ( Q + q) /2      and then the force is    F = k [ (Q + q)/2 ]2 / d2      or     [(Q + q)/2 ]2     =  F d2 /k   =   0.0132 x10-7 Takethe square root of both sides and multiply by 2 and youget             Q + q =    2.30 x10-6 Now...if we measure the charges in microCoulomb we can eliminate theexponents and     Q+ q = 2.30 x10-7         Qq = 0.0132 x10-7 twoequations, two unknowns... solve! Takethe square root of both sides and multiply by 2 and youget             Q + q =    2.30 x10-6 Now...if we measure the charges in microCoulomb we can eliminate theexponents and     Q+ q = 2.30 x10-7         Qq = 0.0132 x10-7 twoequations, two unknowns... solve!

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.