Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 22, Problem 35AP

Three charged particles are aligned along the x axis as shown in Figure P22.35. Find the electric field at (a) the position (2.00 m, 0) and (b) the position (0, 2.00 m).

Figure P22.35

Chapter 22, Problem 35AP, Three charged particles are aligned along the x axis as shown in Figure P22.35. Find the electric

(a)

Expert Solution
Check Mark
To determine

The electric field at the position (2.00 m, 0).

Answer to Problem 35AP

The electric field at the position (2.00 m, 0) is 24.2i^ N/C

Explanation of Solution

Three charges are acting on the same line along the x-axis.

According to Coulomb’s law, write the expression for the electric field created by a charge q

    E=keqr2r^

Here, E is the electric field, ke is the Coulomb’s constant, q is the charge and r is the distance between two charges.

The electric field at a point due to number of charges is resultant of the electric field due to the individual charges.

The figure for the position of point charges is shown below.

Physics for Scientists and Engineers, Chapter 22, Problem 35AP , additional homework tip  1

Figure (1)

Write the expression for the electric field due to charge q1 at P

    E1=keq1r12r^1

Here, E1 is the electric field due to charge q1 and r1 is the distance between the charge q1 and point P.

Substitute 9×109 Nm2/C2 for ke, 4.00 nC for q1 and 2.50 m for r1 in above expression.

    E1=(9×109 Nm2/C2)(4.00 nC)(1C109 nC)(2.50 m)2i=5.76i^N/C

Write the expression for the electric field due to charge q2 at P

    E2=keq2r22r^2

Here, E2 is the electric field due to charge q2 and r2 is the distance between the charge q2 and point P.

Substitute 9×109 Nm2/C2 for ke, 5.00 nC for q2 and 2.00 m for r2 in above expression.

    E2=(9×109 Nm2/C2)(5.00 nC)(1C109 nC)(2.00 m)2i=11.25i^N/C

Write the expression for the electric field due to charge q3 at P

    E3=keq3r32r^3

Here, E3 is the electric field due to charge q3 and r3 is the distance between the charge q3 and point P.

Substitute 9×109 Nm2/C2 for ke, 3.00 nC for q3 and 1.20 m for r3 in above expression.

    E3=(9×109 Nm2/C2)(3.00 nC)(1C109 nC)(1.20 m)2i^=18.75i^N/C

Write the expression for the resultant electric field

    E=E1+E2+E3

Here, E is the resultant electric field.

Conclusion:

Substitute 5.76i^N/C for E1, 11.25i^N/C for E2 and 18.75i^N/C for E3.

    E=(5.76+11.25+18.75 )i^ N/C=24.24i^ N/C

Thus, the electric field at position (2.00 m, 0) is 24.2i^ N/C.

(b)

Expert Solution
Check Mark
To determine

The electric field at the position (0,2.00 m).

Answer to Problem 35AP

The electric field at the position (0,2.00 m) is (4.21)i^+(8.43)j^ N/C

Explanation of Solution

Three charges are acting on the same line along the x-axis.

The figure for the position of point charges is shown below,

Physics for Scientists and Engineers, Chapter 22, Problem 35AP , additional homework tip  2

Figure (2)

According to Pythagoras theorem, the distance between the charge q1 and position Q(0,2.00 m)

    d1=(0.50 m)2+(2.00 m)2=2.06155 m

Here, d1 is the distance between the charge q1 and point Q

According to Pythagoras theorem, the distance between the charge q3 and position Q(0,2.00 m)

    d3=(0.80 m)2+(2.00 m)2=2.154 m

Here, d3 is the distance between the charge q3 and point Q.

Write the expression for the electric field due to charge q1 at Q

    E1=keq1d12d^1

Here, E1 is the electric field due to charge q1.

Substitute 9×109 Nm2/C2 for ke, 4.00 nC for q1 and 2.06155 m for d1,

    E1=(9×109 Nm2/C2)(4.00 nC)(1C109 nC)(2.06155 m)2d^=8.47d^N/C

In component form the electric field can be written as,

    E1=(8.47cosθi^8.47sinθj^)N/C

According to right angle triangle property,

    cosθ=0.502.06=0.2427

And,

    sinθ=2.002.06=0.9708

Substitute 0.2427 for cosθ and 0.9708 for sinθ according to right angle triangle property to get E1,

    E1=(8.47×0.2427)i^(8.47×0.9708)j^ N/C=2.05i^8.22j^ N/C

Write the expression for the electric field due to charge q2 at Q

    E2=keq2d22d^2

Here, E2 is the electric field due to charge q2 and d2 is the distance between the charge q2 and point Q

Substitute 9×109 Nm2/C2 for ke, 5.00 nC for q2 and 2.00 m for d2,

  E2=(9×109 Nm2/C2)(5.00 nC)(1C109 nC)(2.00 m)2d^=11.25j^ N/C

Write the expression for the electric field due to charge q3 at Q

    E3=keq3d32d^3

Here, E3 is the electric field due to charge q3.

Substitute 9×109 Nm2/C2 for ke, 3.00 nC for q3 and 2.154 m for d3,

    E3=(9×109 Nm2/C2)(3.00 nC)(1C109 nC)(2.154 m)2d^=5.819d^ N/C

In component form the electric field can be written as,

    E3=(5.819cosϕ)i^+(5.819sinϕ)j^ N/C

According to right angle triangle property,

    cosϕ=0.802.154=0.371

And,

    sinϕ=2.002.154=0.928

Substitute 0.371 for cosϕ and 0.928 for sinϕ according to right angle triangle property to get E3,

    E3=(5.819×0.371)i^+(5.819×0.928)j^ N/C=2.16i^+5.40j^ N/C

Write the expression for the resultant electric field

    E=E1+E2+E3

Here, E is the resultant electric field

Conclusion:

Substitute 2.05i^8.22j^ N/C for E1, 11.25j^ N/C for E2 and 2.16i^+5.40j^ N/C for E3,

    E=(2.05i^8.22j^+11.25j^2.16i^+5.40j^) N/C=(4.21)i^+(8.43)j^ N/C

Therefore, the electric field at position (0, 2.00 m) is (4.21)i^+(8.43)j^ N/C

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Chapter 22 Solutions

Physics for Scientists and Engineers

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