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 .

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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 two forces on it would not be along the same line and could not sum to zero. Thus the y coordinate of the particle must be zero. The third particle must lie between the other two since otherwise the forces acting on it would be in the same direction and would not sum to zero. Suppose the third particle is a distance x from the particle with charge q, as shown on the diagram to the right. The force acting on it is then given by F0 = 1 4π qq0 4.00qq0 − = 0, 2 x (L − x)2

← x − − L−x − − →← → • • • q q0 4.00q

0

where the positive direction was taken to be toward the right. Solve this equation for x. Canceling common factors yields 1/x2 = 4.00/(L−x)2 and taking the square root yields 1/x = 2.00/(L−x).

The solution is x = 0.333L. The force on q is 1 Fq = 4π

0

### qq0 4.00q 2 + = 0. x2 L2

Solve for q0 : q0 = −4.00qx2 /L2 = −0.444q, where x = 0.333L was used. The force on the particle with charge 4.00q is F4q = 1 4π 1 = 4π 4.00q 2 4.00qq0 1 + = 2 2 L (L − x) 4π 2 2 4.00q 4.00q − = 0. 2 L L2 4.00q 2 4.00(0.444)q 2 + L2 (0.444)L2

0

0

0

With q0 = −0.444q and x = 0.333L, all three charges are in equilibrium.

CHAPTER 21

HINT FOR PROBLEM 7

Calculate the x and y components of the forces of particles 1, 2, and 4 on particle 3. Add the x components to find the x component of the net force and add the y components to find the y component of the net force. Use Coulomb’s law to calculate the magnitude of each force. The charges of particles 1 and 3 have the same sign so the force of 1 is in the negative y direction. The charges of particles 2 and 3 have opposite signs so the force of 2 has positive x and y √ components. The distance between these particles is 2a and the force makes an angle of 45◦ with the positive x direction. The charges of particles 3 and 4 have opposite signs so the force of 4 is in the negative x direction. ans: (a) 0.17 N; (b) −0.046 N

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CHAPTER 21

HINT FOR PROBLEM 19

Divide the shell into concentric shells of infinitesimal thickness dr. The shell with radius r has volume dV = 4πr2 dr and contains charge dq = ρ dV = 4πr2 ρ dr. Integrate this expression. The lower limit is the inner radius of the original shell and the upper limit is its outer radius. ans: 3.8 × 10−8 C

CHAPTER 21

HINT FOR PROBLEM 25

The current is the charge that is intercepted by Earth’s surface per unit time. If N is the number of protons that hit each square meter of the surface per second, then the current is i = N Ae, where A is the area of the surface, given by A = 4πR2. Look up the radius R of Earth in Appendix C. ans: 122 mA