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Chapter 19, Problem 19P

Three point charges are arranged as shown in Figure P19.19. (a) Find the vector electric Field that the 6.00-nC and –3.00-nC charges together create at the origin. (b) Find the vector force on the 5.00-nC charge.

Chapter 19, Problem 19P, Three point charges are arranged as shown in Figure P19.19. (a) Find the vector electric Field that

(a)

Expert Solution
Check Mark
To determine

The electric field at the origin due to charges 6.00nC and 3.00nC.

Answer to Problem 19P

The electric field at the origin due to charges 6.00nC and 3.00nC is E=(5.99×102N/C)i^(2.70×103N/C)j^_.

Explanation of Solution

Figure 1 represents the electric field at the origin due to charges 6.00nC and 3.00nC.

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term, Chapter 19, Problem 19P

Write the expression for the electric field at the origin due to 3.00nC charge.

    E1=ke|q1|r12(j^)        (I)

Here, E1 is the electric field at the origin due to 3.00nC charge, ke is the coulomb’s constant, q1 is the charge, and r1 is the distance from the origin to 3.00nC charge.

Write the expression for the electric field at the origin due to 6.00nC charge.

    E2=ke|q2|r22(i^)        (II)

Here, E1 is the electric field at the origin due to 6.00nC charge, ke is the coulomb’s constant, q1 is the charge, and r1 is the distance from the origin to 6.00nC charge.

Write the expression for net electric field at the origin.

    E=E1+E2        (III)

Conclusion:

Substitute 8.99×109Nm2/C2 for ke, 3.00nC for |q1| 0.100m for r1 in equation (I), to find E1.

    E1=(8.99×109Nm2/C2)|3.00nC×1C109nC|(0.100m)(j^)=(2.70×103N/C)j^        (IV)

Substitute 8.99×109Nm2/C2 for ke, 6.00nC for |q2| 0.300m for r2 in equation (II), to find E2.

    E2=(8.99×109Nm2/C2)|6.00nC×1C109nC|(0.300m)(i^)=(5.99×102N/C)i^        (V)

Use equation (IV) and (V) in equation (III).

    E=(2.70×103N/C)j^(5.99×102N/C)i^

Therefore, the electric field at the origin due to charges 6.00nC and 3.00nC is E=(5.99×102N/C)i^(2.70×103N/C)j^_.

(b)

Expert Solution
Check Mark
To determine

The vector force on the charge 5.00nC.

Answer to Problem 19P

The vector force on the charge 5.00nC is F=(3.00i^13.5j^)μN_.

Explanation of Solution

From subpart (a) the electric field at the origin due to charges 6.00nC and 3.00nC is E=(5.99×102N/C)i^(2.70×103N/C)j^.

Write the expression for vector force on the charge.

    F=qE        (VI)

Here, F is the vector force on the charge, q is the charge, and E is the electric field.

Conclusion:

Substitute 5.00nC for q, and (5.99×102N/C)i^(2.70×103N/C)j^ for E in equation (VI), to find F.

    F=(5.00nC)((5.99×102N/C)i^(2.70×103N/C)j^)N/C=(3.00i^13.5j^)μN

Therefore, the vector force on the charge 5.00nC is F=(3.00i^13.5j^)μN_.

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Chapter 19 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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