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Unformatted text preview: C = −6.2 × 107 N/C ) + (8.99 × 109 N ⋅ m2 /C2 )( 21 × 10−6 C)
2
(9.0 × 10−2 m ) 974 ELECTRIC FORCES AND ELECTRIC FIELDS The minus sign tells us that the net electric field points along the −x axis.
b. The net electric field at x = +6.0 cm (point B) is E = E1 + E2: E = E1 + E2 = k q1
r12 + k q2
r22 (8.99 × 109 N ⋅ m2 /C2 )(8.5 × 10−6 C) + (8.99 × 109 N ⋅ m2 /C2 )( 21 × 10−6 C)
=
2
2
(3.0 × 10−2 m )
(3.0 × 10−2 m )
= +2.9 × 108 N/C
The plus sign tells us that the net electric field points along the +x axis.
______________________________________________________________________________
38. REASONING At every position in space, the
net electric field E is the vector sum of the
external electric field Eext and the electric
field Epoint created by the point charge at the
origin: E = Eext + Epoint. The external electric
field is uniform, which means that it has the
same magnitude Eext = 4500 N/C at all
locations and that it always points in the
positive x direction (see the drawing). The
magnitude Epoint of the electric field due to y
Eext
Epoint
Epoint Epoint Eext the point charge at the origin is given by Epoint = Eext x kq (Equation 18.3), where r is the
r2
distance between the origin and the location where the electric field is to be evaluated, q is
the charge at the origin, and k = 8.99×109 N·m2/C2. All three locations given in the problem
are a distance r = 0.15 m from the origin, so we have that Epoint = kq
r2 (8.99 ×109 N ⋅ m 2 /C2 ) −8.0 ×10−9 C = 3200 N/C
=
( 0.15 m )2 The direction of the electric field Epoint varies from location to location, but because the
charge q is negative, Epoint is always directed towards the origin (see the drawing). Chapter 18 Problems 975 SOLUTION
a. At x = −0.15 m, the electric field of the point charge and the external electric field both
point in the positive x direction (see the drawing), so the magnitude E of the net electric
field is the sum of the magnitudes of the individual electric fields: E = Eext + Epoint = 4500 N/C + 3200 N/C = 7700 N/C
b. At x = +0.15 m, the electric field of the point charge is opposite the external electric field
(see the drawing). Therefore, the magnitude E of the net electric field is the difference
between the magnitudes of the individual electric fields: E = Eext − Epoint = 4500 N/C − 3200 N/C = 1300 N/C
c. At y = +0.15 m, the electric fields Epoint and Eext are perpendicular (see the drawing).
This makes them, in effect, the xcomponent (Eext) and ycomponent (Epoint) of the net
electric field E. The magnitude E of the net electric field, then, is given by the Pythagorean
theorem (Equation 1.7):
2
2
E = Eext + Epoint = ( 4500 N/C )2 + ( 3200 N/C )2 = 5500 N/C 39. SSM REASONING Since the charged droplet (charge = q) is suspended motionless in
the electric field E, the net force on the droplet must be zero. There are two forces that act
on the droplet, the force of gravity W = m g , and the electric force F = qE due to the electric
field. Since the net force on the droplet is zero, we conclude that mg = q E . We can use
this reasoning to determine the sign and the magnitude of the charge on the droplet.
SOLUTION
a. Since the net force on the droplet is zero, and the weight of magnitude W
points downward, the electric force of magnitude F = q E must point
upward. Since the electric field points upward, the excess charge on the
droplet must be positive in order for the force F to point upward. F mg b. Using the expression mg = q E , we find that the magnitude of the excess charge on the
droplet is
mg (3.50 ×10 –9 kg)(9.80 m/s 2 )
q=
=
= 4.04 ×10 –12 C
E
8480 N/C 976 ELECTRIC FORCES AND ELECTRIC FIELDS The charge on a proton is 1.60 × 10–19 C, so the excess number of protons is
1 proton
( 4.04 ×10–12 C ) 1.60 ×10–19 C = 2.53 ×107 protons ______________________________________________________________________________
40. REASONING Since the proton and the electron have the same charge magnitude e, the
electric force that each experiences has the same magnitude. The directions are different,
however. The proton, being positive, experiences a force in the same direction as the electric
field (due east). The electron, being negative, experiences a force in the opposite direction
(due west).
Newton’s second law indicates that the direction of the acceleration is the same as the
direction of the net force, which, in this case, is the electric force. The proton’s acceleration
is in the same direction (due east) as the electric field. The electron’s acceleration is in the
opposite direction (due west) as the electric field.
Newton’s second law indicates that the magnitude of the acceleration is equal to the
magnitude of the electric force divided by the mass. Although the proton and electron
experience the same force magnitude, they have different masses. Thus, they have
accelerations of different magnitudes. SOLUTION According to Newton’s second law, Equation 4.2, the acceleration a of an
object is equal to the net force divided by the object’s mass m. In this situ...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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