Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 23, Problem 23.18P

Particle A of charge 3.00 × 10−4 C is at the origin, particle B of charge −6.00 × 10−4 C is at (4.00 m, 0), and particle C of charge 1.00 × 10−4 C is at (0, 3.00 m). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C? (b) What is the y component of the force exerted by A on C? (c) Find the magnitude of the force exerted by B on C. (d) Calculate the x component of the force exerted by B on C. (e) Calculate the y component of the force exerted by B on C. (f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. (g) Similarly, find the y component of the resultant force vector acting on C. (h) Find the magnitude and direction of the resultant electric force acting on C.

(a)

Expert Solution
Check Mark
To determine

The x component of the electric force exerted by A on C.

Answer to Problem 23.18P

The x component of the electric force exerted by A on C is 0.

Explanation of Solution

The charge of particle A is 3.00×104C, charge of particle B is 6.00×104C, and charge of particle C is 1.00×104C, location of particle A is origin, location of particle B is (4.00m,0) and  location of particle C is (0,3.00m).

The diagram for the given condition is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 23, Problem 23.18P

Figure 1

Write the formula to calculate the electrical force

    FACx=keq1q2r2

Here, ke is the constant, q1 is the charge on particle A, q2 is the charge on particle C and r is the distance between the particles A and C on x axis.

The particle A is at origin that is the abscissa and the ordinate both are 0 and the particle C is at the distance of 3.00m from the y axis and the abscissa is 0.

The distance from the x axis is 0 for both the particles A and C. So, no net force will act on the particle A by C in the x-direction

Thus, the x component of the electric force exerted by A on C

    FACx=0

Conclusion:

Therefore, the x component of the electric force exerted by A on C is 0.

(b)

Expert Solution
Check Mark
To determine

The y component of the force exerted by A on C.

Answer to Problem 23.18P

The y component of the force exerted by A on C is 30N.

Explanation of Solution

Write the formula to calculate the electrical force

    FACy=keq1q2r2

Substitute 3.00×104C for q1, 1.00×104C for q2, 3.00m to calculate FACy as,

    FACy=(9×109Nm2/C2)(3.00×104C)(1.00×104C)(3.00m)2=30.0N

Conclusion:

Therefore, the y component of the force exerted by A on C is 30N.

(c)

Expert Solution
Check Mark
To determine

The magnitude of the force exerted by B on C.

Answer to Problem 23.18P

The magnitude of the force exerted by B on C is 21.6N.

Explanation of Solution

By Pythagoras theorem, write the expression distance between B and C

    (BC)2=(AC)2+(AB)2=(3m)2+(4m)2BC=(16+9)m2=5m

Write the formula to calculate the electrical force

    FBC=keq1q2r2

Here, ke is the constant, q1 is the charge on particle B, q2 is the charge on particle C and r is the distance between the particles B and C.

Substitute 6.00×104C for q1, 1.00×104C for q2, 5.00m to find FBC as,

    FBC=keq1q2r2=(9×109Nm2/C2)(6.00×104C)(1.00×104C)(5.00m)2=21.6N|FBC|=21.6N

Conclusion:

Therefore, the magnitude of the force exerted by B on C 21.6N.

(d)

Expert Solution
Check Mark
To determine

The x component of the force exerted by B on C.

Answer to Problem 23.18P

The x component of the force exerted by B on C is 17.3N.

Explanation of Solution

From part (c), the magnitude of the force exerted by B on C is 21.6N.

Resolve the side BC into horizontal and vertical components as- horizontal component is |FBC|cosα and vertical component is |FBC|sinα

From Figure I

    cosα=45

Here, α is the angle between the x-axis and the line joining the charges B and C.

Write the formula to calculate the x component of force by B on C

    FxBC=|FBC|cosα

Here, |FBC| is the force exerted by B on C and cosα is the horizontal component of BC.

Substitute 21.6N for |FBC| and 45 for cosα to calculate FxBC as,

    FxBC=|FBC|cosα=(21.6N)(45)=17.28N17.3N

Conclusion:

Therefore, the x component of the force exerted by B on C is 17.3N.

(e)

Expert Solution
Check Mark
To determine

The y component of the force exerted by B on C.

Answer to Problem 23.18P

The y component of the force exerted by B on C is 13.0N.

Explanation of Solution

From part (c), the magnitude of the force exerted by B on C is 21.6N.

Resolve the side BC into horizontal and vertical components as-the horizontal component is |FBC|cosα and vertical component is |FBC|sinα

From Figure I,

    sinα=35

Write the formula to calculate the y component of force by B C

    FyBC=|FBC|sinα

Here, |FBC| is the force exerted by B on C and sinα is the vertical component of BC.

Substitute 21.6N for |FBC| and 35 for sinα in the above formula as,

    FyBC=|FBC|sinα=(21.6N)(35)=12.96N13.0N

Conclusion:

Therefore, the y component of the force exerted by B on C is 13.0N.

(f)

Expert Solution
Check Mark
To determine

The resultant x component of the electric force acting on C.

Answer to Problem 23.18P

The resultant x component of the electric force acting on C is 17.3N

Explanation of Solution

From part (a), the x component of the force exerted by A on C

    |FxAC|=0

From part (d), the x component of the force exerted by B on C

    |FxBC|=17.28N

Write the formula to calculate the resultant force acting on the particle C

    FxC=FxAC+FxBC

Here, FxC is the sum of x component of the force acting on C, FxAC is the x component of the force exerted by A on C and FxBC is the x component of the force exerted by B on C.

Substitute 0 for |FxAC| and 17.3N for |FxBC| to calculate FxC as,

    FxC=FxAC+FxBC=0+17.3N=17.3N

Conclusion:

Therefore, the resultant x component of the electric force acting on C is 17.3N.

(g)

Expert Solution
Check Mark
To determine

The resultant y component of the electric force acting on C.

Answer to Problem 23.18P

The resultant y component of the electric force acting on C is 17.0N

Explanation of Solution

From part (b), the y component of the force exerted by A on C

    |FyAC|=30N

From part (e), the y component of the force exerted by B on C

    |FyBC|=17.28N

Write the formula to calculate the resultant force acting on the particle C

    FyC=FyAC+FyBC

Here, FyC is the sum of y component of the force acting on C, FyAC is the y component of the force exerted by A on C and FyBC is the y component of the force exerted by B on C.

Substitute 30N for |FyAC| and 13.0N for |FyBC| to calculate FyC as,

    FyC=FyAC+FyBC=30N13.0N=17.0N

Conclusion:

Therefore, the resultant y component of the electric force acting on C is 17.0N.

(h)

Expert Solution
Check Mark
To determine

The magnitude and direction of the resultant electric force acting on C

Answer to Problem 23.18P

The magnitude and direction of the resultant electric force acting on C is 24.3N and at an angle 44.5° with the x-axis.

Explanation of Solution

From part (g), the resultant y component of the electric force acting on C is 42.96N.

    |FyC|=42.96N

From part (f), the resultant x component of the electric force acting on C

    |FxC|=17.28N

Write the formula to calculate the resultant force acting on the particle C

    FC=F2yC+F2xC

Here, FC is the resultant force acting on C, FyC is the resultant y component of the force exerted by A on C and FxC is the x component of the force exerted by B on C.

Substitute 17.0N for |FyC| and 17.3N for |FxC| to calculate FC  as,

    FC=F2yC+F2xC=(17.0N)2+(17.3N)2=24.3N

Write the formula to calculate the direction of the resultant force acting on C

    θ=tan1(FyCFxC)

Here, Fy is the resultant y component force acting on the particle C and Fx is the resultant x component force acting on the particle C.

Substitute 17.0N for FyC and 17.3N for FxC to calculate θ

    θ=tan1(FyCFxC)=tan1(17.0N17.3N)=44.5°

The direction of the resultant force is counterclockwise from +x axis.

Conclusion:

Therefore, the magnitude and direction of the resultant electric force acting on C is 24.3N and at an angle 44.5° with the x-axis.

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Chapter 23 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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