Ejercicios Fisica

PHY2049

R. D. Field

Chapter 22 Solutions
Problem 1:
A +15 microC charge is located 40 cm from a +3.0 microC charge. The magnitude of the electrostatic force on the larger charge and on the smaller charge (in N) is, respectively, Answer: 2.5, 2.5 Solution: The magnitiude of the electrostatic for is given by,

KQ1Q2 (8.99 × 109 Nm 2 / C 2 )(15µC )(3µC ) = = 2.53N . F= r2 (40cm) 2Remember the force on 1 due to 2 is equal and opposite to the force on 2 due to 1.

Problem 2:
Two point particles have charges q1 and q2 and are separated by a distance d. Particle q2 experiences an electrostatic force of 12 milliN due to particle q1. If the charges of both particles are doubled and if the distance between them is doubled, what is the magnitude of the electrostatic force between them(in milliN)? Answer: 12 Solution: The magnitude of the initial electristatic force is

Fi = Ff =

Kq1q2 = 12mN . d2

The magnitude of the final electrostatic force is

K (2q1 )(2q2 ) = Fi = 12mN . ( 2d ) 2
+Q r d x F1 +Q q F2 θ

Problem 3:
Two identical point charges +Q are located on the y-axis at y=+d/2 and y=-d/2, as shown in the Figure. A third charge q is placed on the x-axis. Atwhat distance from the origin is the net force on q a maximum?
Solutions Chapter 22

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R. D. Field

d /( 2 2 ) Answer: Solution: The net force on q is the superposition of the forces from each of the two charges as follows:

r KQq ˆ ˆ (cos θ x − sin θ y ) F1 = 2 r r KQq ˆ ˆ (cos θ x + sin θ y ) F2 = 2 r r r r 2 KQq ˆ cos θ x F = F1 + F 2 = 2 r 2 KQqx ˆ = x ( x 2 +(d / 2) 2 )3 / 2

where I used r2 = x2 + (d/2)2 and cosθ = x/r. To find the maximum we θ take the derivative of F and set it equal to zero as follows:

  1 3x 2 dF = 2 KQq 2 − 2  ( x + (d / 2) 2 )3 / 2 ( x + (d / 2) 2 )5 / 2   dx   2 KQq (d / 2) 2 − 2 x 2 = =0 ( x 2 + (d / 2) 2 )5 / 2
Solving for x gives x = d /(2 2 ) .

(

)

Problem 4:
Two 2.0 gram balls hang from lightweightinsulating threads 50 cm long from a common support point as shown in the Figure. When equal charges Q are place on each ball they are repelled, each making an angle of 10 degrees with the vertical. What is the magnitude of Q, in microC? Answer: 0.11 Solution: Requiring the x and y components of the force to vanish yields
θ
L

QE Q

T Q r mg

Solutions

Chapter 22

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R. D. Field

KQ2 T sin θ = QE = 2 r . T cosθ = mg
Eliminating the tension T and solving for Q2 gives

mgr 2 tan θ 4 mgL2 sin 2 θ tan θ = Q = K K
2

and
Q= 4 ( 2 × 10 −3 kg )( 9.8m / s 2 )( 0.5m) 2 sin 2 10 0 tan 10 0 = 0.11µC . 8.99 × 10 9 Nm 2 / C 2
-2Q Q

Problem 5:
Four point charges q, Q, -2Q, and Q form a square with sides of length L as shown in the Figure. What is themagnitude of the resulting electrostatic force on the charge q due to the other three charges? Answer: 0.41KqQ/L2 Solution: The electrostatic force on q is the vector superposition of the following three forces,

L

3 1
Q L q

tot 2

r KqQ ˆ F1 = 2 x L r KqQ ˆ F2 = − 2 y L r 2 KqQ  1 1  ˆ ˆ F3 = x+ y − ( 2 L) 2  2 2  r KqQ  1  ˆ ˆ Ftot = 2 1 − ( x − y ) . L  2

and the netforce is thus,

The magnitude is given by

r KqQ Ftot = 2 L

(

2 − 1 = 0.41

)

KqQ . L2

Solutions

Chapter 22

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R. D. Field

Problem 6:

L=1 m Two 2.0 gram charged balls hang from lightweight insulating threads 1 m long from a θ T common support point as shown in the Figure. Q2 If one of the balls has a charge of 0.01 microC Q1 x = 15 cm and if theballs are separated by a distance of 15 cm, what is the charge on the other ball (in mg microC)? Answer: 0.37 Solution: The conditions for static equilibrium on ball Q2 is

θ

L=1m

KQ1Q2/x

2

T sin θ =

KQ1Q2 x2 T cosθ = mg

and dividing the 1st equation by the second yields

tan θ =

KQ1Q2 x 2 mg . Solving for the product of the two charges gives

( 0 .15 m ) 3 ( 2 × 10 −3...