EXAMPLE 15.5 ELECTRIC FIELD DUE TO TWO POINT CHARGES GOAL Use the superposition principle to calculate the electric field due to two point charges. PROBLEM Charge q ==== 7.00 μC is at the origin, and charge q2 = -5.00 μC is on the x-axis, 0.300 m from the origin (Fig. 15.12). (a) Find the magnitude and direction of the electric field at point P, which has coordinates (0, 0.400) m. (b) Find the force on a charge of 2.00 x 10-8 C placed at P. STRATEGY Follow the problem-solving strategy, finding the electric field at point P due to each indi- vidual charge in terms of x- and y-components, then adding the components of each type to get the x- and y-components of the resultant electric field at P. The magnitude of the force in part (b) can be found by simply multiplying the magnitude of the electric field by the charge. Figure 15.12 (Example 15.5) The resultant electric field E at P equals the vector sum Ē₁ + E₂, where E, is the field due to the positive charge q and E, is the field due to the negative charge 92. E₁ E P 102 0.400 m 0.500 m 0.300 m 91 92 I EXERCISE 15.5 (a) Place a charge of -7.00 μC at point P and find the magnitude and direction of the electric field at the location of 92 due to qi and the charge at P. (b) Find the magnitude and direction of the force on q2. ANSWERS (a) 5.84 x 105 N/C, & = 20.2° (b) F= 2.92 N, = 200°.
EXAMPLE 15.5 ELECTRIC FIELD DUE TO TWO POINT CHARGES GOAL Use the superposition principle to calculate the electric field due to two point charges. PROBLEM Charge q ==== 7.00 μC is at the origin, and charge q2 = -5.00 μC is on the x-axis, 0.300 m from the origin (Fig. 15.12). (a) Find the magnitude and direction of the electric field at point P, which has coordinates (0, 0.400) m. (b) Find the force on a charge of 2.00 x 10-8 C placed at P. STRATEGY Follow the problem-solving strategy, finding the electric field at point P due to each indi- vidual charge in terms of x- and y-components, then adding the components of each type to get the x- and y-components of the resultant electric field at P. The magnitude of the force in part (b) can be found by simply multiplying the magnitude of the electric field by the charge. Figure 15.12 (Example 15.5) The resultant electric field E at P equals the vector sum Ē₁ + E₂, where E, is the field due to the positive charge q and E, is the field due to the negative charge 92. E₁ E P 102 0.400 m 0.500 m 0.300 m 91 92 I EXERCISE 15.5 (a) Place a charge of -7.00 μC at point P and find the magnitude and direction of the electric field at the location of 92 due to qi and the charge at P. (b) Find the magnitude and direction of the force on q2. ANSWERS (a) 5.84 x 105 N/C, & = 20.2° (b) F= 2.92 N, = 200°.
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter19: Electric Forces And Electric Fields
Section: Chapter Questions
Problem 15P
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Transcribed Image Text:EXAMPLE 15.5 ELECTRIC FIELD DUE TO TWO POINT CHARGES
GOAL Use the superposition principle to calculate
the electric field due to two point charges.
PROBLEM Charge q ==== 7.00 μC is at the origin,
and charge q2 = -5.00 μC is on the x-axis, 0.300 m
from the origin (Fig. 15.12). (a) Find the magnitude
and direction of the electric field at point P, which
has coordinates (0, 0.400) m. (b) Find the force on a
charge of 2.00 x 10-8 C placed at P.
STRATEGY Follow the problem-solving strategy,
finding the electric field at point P due to each indi-
vidual charge in terms of x- and y-components, then
adding the components of each type to get the x- and
y-components of the resultant electric field at P. The
magnitude of the force in part (b) can be found by
simply multiplying the magnitude of the electric field
by the charge.
Figure 15.12 (Example 15.5)
The resultant electric field E at P
equals the vector sum Ē₁ + E₂, where
E, is the field due to the positive
charge q and E, is the field due to
the negative charge 92.
E₁
E
P
102
0.400 m
0.500 m
0.300 m
91
92
I
EXERCISE 15.5 (a) Place a charge of -7.00 μC at point P and find the magnitude and direction of the electric field at the
location of 92 due to qi
and the charge at P. (b) Find the magnitude and direction of the force on q2.
ANSWERS (a) 5.84 x 105 N/C, & = 20.2° (b) F= 2.92 N, = 200°.
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