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CHAPTER 21

SOLUTION FOR PROBLEM 9

Assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be used. Let q1 and q2 be the original charges and choose the coordinate system so the force on q2 is positive if it is repelled by q1 . Take the distance between the charges to be r. Then the force on q2 is Fa = − 1 4π

0

q1 q2 . r2

The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, have the same charge. Since charge is conserved, the total charge is the same as it was originally. This means the charge on each sphere is (q1 + q2 )/2. The force is now one of repulsion and is given by Fb = 1 4π

0

(q1 + q2 )2 . 4r 2

Solve the two force equations simultaneously for q and q2 . The first gives 1 q1 q2 = −4π 0 r Fa = − and the second gives q1 + q2 = 2r Thus q2 = and

2

(0.500 m)2 (0.108 N) 8.99 × 109 N·

2 m2 /C

= −3.00 × 10−12 C2

4π 0 Fb = 2(0.500 m)

0.0360 N 8.99 × 109 N ·

2 m2 /C

= 2.00 × 10−6 C .

−(3.00 × 10−12 C2 ) q1

3.00 × 10−12 C2 = 2.00 × 10−6 C . q1 Multiply by q1 to obtain the quadratic equation q1 −

2 q1 − (2.00 × 10−6 C)q1 − 3.00 × 10−12 C2 = 0 .

The solutions are q1 = 2.00 × 10−6 C ± (−2.00 × 10−6 C)2 + 4(3.00 × 10−12 C2 ) . 2

If the positive sign is used, q1 = 3.00 × 10−6 C and if the negative sign is used, q1 = −1.00 × 10−6 C. Use q2 = (−3.00 × 10−12 )/q1 to calculate q2 . If q1 = 3.00 × 10−6 C, then q2 = −1.00 × 10−6 C and if q1 = −1.00 × 10−6 C, then q2 = 3.00 × 10−6 C. Since the spheres are −6 identical, the solutions are essentially the same: one sphere originally had charge −1.00 × 10 C and the other had charge +3.00 × 10−6 C.

CHAPTER 21

SOLUTION FOR PROBLEM 17

If the system of three particles is to be in equilibrium, the force on each particle must be zero. Let the charge on the third particle be q0 . The third particle must lie on the x axis since otherwise the...

SOLUTION FOR PROBLEM 9

Assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be used. Let q1 and q2 be the original charges and choose the coordinate system so the force on q2 is positive if it is repelled by q1 . Take the distance between the charges to be r. Then the force on q2 is Fa = − 1 4π

0

q1 q2 . r2

The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, have the same charge. Since charge is conserved, the total charge is the same as it was originally. This means the charge on each sphere is (q1 + q2 )/2. The force is now one of repulsion and is given by Fb = 1 4π

0

(q1 + q2 )2 . 4r 2

Solve the two force equations simultaneously for q and q2 . The first gives 1 q1 q2 = −4π 0 r Fa = − and the second gives q1 + q2 = 2r Thus q2 = and

2

(0.500 m)2 (0.108 N) 8.99 × 109 N·

2 m2 /C

= −3.00 × 10−12 C2

4π 0 Fb = 2(0.500 m)

0.0360 N 8.99 × 109 N ·

2 m2 /C

= 2.00 × 10−6 C .

−(3.00 × 10−12 C2 ) q1

3.00 × 10−12 C2 = 2.00 × 10−6 C . q1 Multiply by q1 to obtain the quadratic equation q1 −

2 q1 − (2.00 × 10−6 C)q1 − 3.00 × 10−12 C2 = 0 .

The solutions are q1 = 2.00 × 10−6 C ± (−2.00 × 10−6 C)2 + 4(3.00 × 10−12 C2 ) . 2

If the positive sign is used, q1 = 3.00 × 10−6 C and if the negative sign is used, q1 = −1.00 × 10−6 C. Use q2 = (−3.00 × 10−12 )/q1 to calculate q2 . If q1 = 3.00 × 10−6 C, then q2 = −1.00 × 10−6 C and if q1 = −1.00 × 10−6 C, then q2 = 3.00 × 10−6 C. Since the spheres are −6 identical, the solutions are essentially the same: one sphere originally had charge −1.00 × 10 C and the other had charge +3.00 × 10−6 C.

CHAPTER 21

SOLUTION FOR PROBLEM 17

If the system of three particles is to be in equilibrium, the force on each particle must be zero. Let the charge on the third particle be q0 . The third particle must lie on the x axis since otherwise the...

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