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SP212 - Homework 6 Solutions LCDR Brett Rabbit Williams, USN

1. Each of two small non-conducting spheres is charged positively, the combined charge being 40 C.When the two spheres are 50 cm apart, each sphere is repelled by the other by a force of magnitude

2.0 N. Determine the magnitude of the smaller of the two charges.(a) 1.4 C

(b) 1.1 C

(c) 2.0 C

(d) 3.3 C

(e) 17 C

Solution

The product of the charges must satisfy

2 = k q1q2

(0.5)2. (1.1)

Since the charges sum to 40 C, we may make the substitution

q2 = 4105 q1, (1.2)and solve for q1. This yields the two roots

q1 ={3.9105, 1.4106}, (1.3)the smaller of which may be taken as the magnitude of the smaller charge.

2. A particle (m = 50 g, q = 5.0 C) is released from rest when it is 50 cm from a second particle(Q=20 C). Determine the magnitude of the initial acceleration of the 50-g particle.

(a) 54 m/s2

(b) 90 m/s2

(c) 72 m/s2

(d) 65 m/s2

(e) 36 m/s2

Solution

The only force acting on the particle is the electric force, thus

F =ma = q1E. (2.1)

Solving for a, we have

a=|q1|E

m

(2.2)

= k|q1||q2|

r2m (2.3)

=(9109)(5106)(2105)

(0.5)2(0.05) (2.4)

= 72. (2.5)

Sections 1125 & 2145 Page 1 of 5 Spring 2009


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3. A +20-nC point charge is placed on the x-axis at x= 2.0 m, and a25-nC point charge is placed onthe y -axis at y =3.0 m. What is the direction of the electric field at the origin?

(a) 209

(b) 61

(c) 29

(d) 241

(e) 151

Solution

By inspection, we conclude that the resulting field vector must lie in the third quadrant, and the angleit makes with the x-axis is

= tan1Ey

Ex

(3.1)

= tan1

k|q|y2

k|q+|x2

(3.2)

= tan1

2.51089

2.01084

(3.3)

= tan1

5

9

(3.4)

= 29. (3.5)

For a vector in the third quadrant, this corresponds to an angle of 209

from the positive x-axis.

4. A point charge (5.0 pC) is located at the center of a spherical surface (radius = 2.0 cm), and a chargeof 3.0 pC is spread uniformly upon this surface. Determine the magnitude of the electric field 1.0 cmfrom the point charge.

(a) 0.72 kN/C

(b) 0.45 kN/C

(c) 0.63 kN/C

(d) 0.90 kN/C

(e) 0.18 kN/C

Solution

The field inside the spherical shell is due only to the point charge at the center, given by

E= ke|q|r2

(4.1)

= (9109) 51012

(0.01)2 (4.2)

= 450. (4.3)



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5. Which one of the following is not an expression for electric charge?

(a) dV

(b) dA

(c) dl

(d)E dA

(e) 0E dA

Solution

The expression

E dA (5.1)has units of Nm2/C, and so cannot represent electric charge.

6. The electric field just outside the surface of a hollow conducting sphere of radius 20 cm has a magnitudeof 500 N/C and is directed outward. A charge Q is introduced into the center of the sphere and itis noted that the electric field is still directed outward but has decreased to 100 N/C. What is themagnitude of the charge Q?

(a) 1.5 nC

(b) 1.8 nC

(c) 1.3 nC

(d) 1.1 nC

(e) 2.7 nC

Solution

By superposition we have

100 =Esphere+EQ (6.1)

= 500 +k Q

(0.2)2. (6.2)

Solving for Q, we have

Q=

400(0.2)2

k

=

1.8

109. (6.3)

7. An electron (m = 9.11031 kg, q=1.61019 C) starts from rest at point A and has a speedof 5.0106 m/s at point B. Only electric forces act on it during this motion. Determine the electricpotential difference VAVB.

(a) 71 V



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