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Unformatted text preview: e magnitude q of the
charge. Thus, the x component Ex of the net electric field is proportional to
5.0 µC (2.0 µC + 3.0 µC). Since only one of the charges produces an electric field in the y
direction, the y component Ey of the net electric field is proportional to the magnitude of this charge, or 5.0 µC. Thus, the x and y components are equal, as indicated at the righthand
side of the following drawing, where the net electric field E is also shown.
−5.0 µ C E− 5 Ey
E+2 +2.0 µ C O E− 3 E −3.0 µ C Ex Part (b) of the drawing given in the text. Using the same arguments as earlier, we find
that the electric fields produced by the four charges are shown at the lefthand side of the
following drawing. These fields also produce the same net electric field E as before, as
indicated at the righthand side of the following drawing. Chapter 18 Problems E+6 971 +1.0 µ C +4.0 µ C E Ey E− 1
−1.0 µ C Ex E+4
E+1
+6.0 µ C SOLUTION
Part (a) of the drawing given in the text. The net electric field in the x direction is Ex (8.99 × 109 N ⋅ m2 /C2 ) ( 2.0 × 10−6 C ) + (8.99 × 109 N ⋅ m2 /C2 ) (3.0 × 10−6 C )
=
( 0.061 m )2 ( 0.061 m )2 = 1.2 × 107 N /C
The net electric field in the y direction is Ey (8.99 × 109 N ⋅ m2 /C2 ) ( 5.0 × 10−6 C ) = 1.2 × 107 N /C
=
( 0.061 m )2 The magnitude of the net electric field is
2
2
E = Ex + E y = (1.2 × 107 N /C ) + (1.2 × 107 N /C )
2 2 = 1.7 × 107 N /C Part (b) of the drawing given in the text. The magnitude of the net electric field is the same as determined for part (a); E = 1.7 × 107 N/C .
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35. REASONING AND SOLUTION
a. In order for the field to be zero, the point cannot be between the two charges. Instead, it
must be located on the line between the two charges on the side of the positive charge and
away from the negative charge. If x is the distance from the positive charge to the point in
question, then the negative charge is at a distance (3.0 m + x) meters from this point. For
the field to be zero here we have
k q− ( 3.0 m + x )2 = k q+
x2 or q− ( 3.0 m + x )2 = q+
x2 972 ELECTRIC FORCES AND ELECTRIC FIELDS Solving for the ratio of the charge magnitudes gives 16.0 µ C ( 3.0 m + x )
=
=
4.0 µ C
q+
x2
q− ( 3.0 m + x )2
4.0 = 2 or x2 Suppressing the units for convenience and rearranging this result gives 4.0x 2 = ( 3.0 + x ) 2 or 4.0x 2 = 9.0 + 6.0 x + x 2 or 3x 2 − 6.0 x − 9.0 = 0 Solving this quadratic equation for x with the aid of the quadratic formula (see Appendix
C.4) shows that
x = 3.0 m or x = −1.0 m
We choose the positive value for x, since the negative value would locate the zerofield spot
between the two charges, where it can not be (see above). Thus, we have x = 3.0 m .
b. Since the field is zero at this point, the force acting on a charge at that point is 0 N .
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36. REASONING
a. The magnitude E of the electric field is given by E = σ / ε 0 (Equation 18.4), where σ is
the charge density (or charge per unit area) and ε 0 is the permittivity of free space.
b. The magnitude F of the electric force that would be exerted on a K+ ion placed inside the
membrane is the product of the magnitude q0 of the charge and the magnitude E of the
electric field (see Equation 18.2), or F = q0 E . SOLUTION
a. The magnitude of the electric field is E= σ
7.1× 10−6 C/m 2
=
= 8.0 ×105 N/C
−12 2
2
ε 0 8.85 ×10 C / N ⋅ m ( ) + b. The magnitude F of the force exerted on a K ion (q0 = +e) is
F = q0 E = e E = 1.60 × 10 −19 C ( 8.0 × 105 N/C ) = 1.3 × 10 −13 N ______________________________________________________________________________ Chapter 18 Problems 37. 973 SSM REASONING
a. The drawing shows the two point charges q1 and q2. Point A is located at x = 0 cm, and
point B is at x = +6.0 cm. A E1 3.0 cm 3.0 cm B 3.0 cm
q2 q1 E2 Since q1 is positive, the electric field points away from it. At point A, the electric field E1 points to the left, in the −x direction. Since q2 is negative, the electric field points toward it.
At point A, the electric field E2 points to the right, in the +x direction. The net electric field
is E = −E1 + E2. We can use Equation 18.3, E = k q / r 2 , to find the magnitude of the
electric field due to each point charge.
b. The drawing shows the electric fields produced by the charges q1 and q2 at point B,
which is located at x = +6.0 cm. A 3.0 cm 3.0 cm
q1 B 3.0 cm q2
E1
E2 Since q1 is positive, the electric field points away from it. At point B, the electric field
points to the right, in the +x direction. Since q2 is negative, the electric field points toward
it. At point B, the electric field points to the right, in the +x direction. The net electric field
is E = +E1 + E2.
SOLUTION
a. The net electric field at the origin (point A) is E = −E1 + E2: E = − E1 + E2 = = ( −k q1
r12 + k q2
r22 )(
2
(3.0 × 10−2 m ) − 8.99 × 109 N ⋅ m 2 /C2 8.5 × 10−6...
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 Spring '13
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 Physics, The Lottery

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