GSO003 Problem 3
Problem Statement
Measuring Unknown Charges with a Coulomb Torsion Balance
Problem Statement
A physics lab experiment uses a Coulomb torsion balance to determine the charges on two small metal spheres A and B. The spheres are identical conducting spheres of the same size.
From prior qualitative tests (triboelectric series, etc.), it is known that sphere A carries positive charge +QA (QA>0) and sphere B carries negative charge −QB (QB>0), with QA>QB.
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Spheres A and B are placed with their centers a distance r apart. The twist angle of the torsion wire shows an attractive force of magnitude F1 acting between them.
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Using an insulating rod, spheres A and B are brought into contact once. After charge redistribution is complete, the spheres are separated again to center-to-center distance r.
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Now a repulsive force of magnitude F2 is observed between the two spheres.
No charge leaks to the surroundings. Each sphere is small enough relative to r to be treated as a point charge.
From these observations, find the initial charge magnitudes QA and QB.
Constraints
- Coulomb's law constant: k=9.0×109 N⋅m2/C2
- Center-to-center distance: r=3.0×10−2 m
- Attractive force before contact: F1=2.4×10−4 N
- Repulsive force after contact: F2=1.0×10−5 N
Input Format
The values QA and QB can each be written as x×10−9 C and y×10−9 C respectively (where x,y are positive integers).
Compute 10x+y and give the answer as a positive integer.
Solution
1. Before Contact: Attractive Force F1
F1=kr2QAQB QAQB=kF1r2=9.0×109(2.4×10−4)×(3.0×10−2)2=9.0×1092.4×10−4×9.0×10−4=24×10−18 C22. After Contact: Repulsive Force F2
When two identical conducting spheres touch, the total charge QA−QB is shared equally. Each sphere gets:
Q′=2QA−QB F2=kr2(Q′)2=k4r2(QA−QB)2 (QA−QB)2=k4F2r2=9.0×1094×(1.0×10−5)×(9.0×10−4)=4.0×10−18 C2Since QA>QB:
QA−QB=2.0×10−9 C3. Solving for QA and QB
Using (QA+QB)2=(QA−QB)2+4QAQB:
(QA+QB)2=4.0×10−18+96×10−18=100×10−18 QA+QB=10×10−9 CSolving the system:
QA=6.0×10−9 C,QB=4.0×10−9 CSo x=6, y=4:
10x+y=60+4=64Answer: 64