PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 16, Problem 22P
To determine

The electric force on the +1.0μC charge due to the other two charges.

Expert Solution & Answer
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Answer to Problem 22P

The electric force on the +1.0μC charge due to the other two charges is 1.2N at 28° below the negative x-axis_.

Explanation of Solution

The arrangement of the charges is given in the figure. Let us represent +1.0μC as q1, 0.60μC as q2, and +0.80μC as q3.

Let us consider as the charges q1 and q2 be in the x-axis, with q1 at the origin. Let i^ and j^ represents the unit vectors along x and y directions.

The use of Pythagoras theorem to the right triangle arrangement of the charges, the distance between the charges q1 and q2 is 6.0cm. Hence the distance between q1 and q2 is 6.0cm (0.06m), and that between q1 and q3 is 10.0cm (0.100m).

According to superposition principle, the force on charge q1 is the sum of forces due to q2 and q3.

Write the expression for the force on charge q1 due to q2.

F12=kq1(q2r123r12) (I)

Here, F12 is the force of charge q1 due to q2, k is a constant (k=8.988×109Nm2/C2), r12 is the distance between q1 and q2, and r12 is the position vector of q2 from the origin.

Write the expression for the force on charge q1 due to q3.

F13=kq1(q3r133r13) (II)

Here, F13 is the force of charge q1 due to q3, r13 is the distance between q1 and q3, and r13 is the position vector of q3 from the origin.

Write the expression for the total force on the charge q1.

Ftotal=F12+F13 (III)

Write the expression for the position vectors r12 and r13 from the figure.

r12=0.060i^+0.00j^ (IV)

r13=0.060i^+0.080j^ (V)

The magnitude of position vectors r12 and r13 is obtained from figure as, r12=0.060m, and r13=0.100m respectively.

Use equation (IV) in (I).

F12=kq1q2r123(0.060i^+0.00j^) (VI)

Use equation (V) in (II).

F13=kq1q3r133(0.060i^+0.080j^) (VII)

Use equation (VI) and (VII) in (III).

Ftotal=kq1q2r123(0.060i^+0.00j^)+kq1q3r133(0.060i^+0.080j^)=kq1[q2r123(0.060i^+0.00j^)+q3r133(0.060i^+0.080j^)]=[kq1(q2r123+q3r133)(0.060)]i^+[kq1q3r133(0.080)]j^ (VIII)

Write the expression for the magnitude of the total force.

Ftotal=Fx2+Fy2 (IX)

Here, Fx and Fy are the x and y components of the total force respectively.

Let θ be the angle that the resultant force makes below the negative x-axis. It is given by the expression,

θ=tan1(FyFx) (X)

Conclusion:

Substitute 8.988×109Nm2/C2 for k, +1.0μC for q1, 0.60μC for q2, +0.80μC for q3, 0.060m for r12, and 0.100m for r13 in equation (VIII) to find Ftotal.

Ftotal=[(8.988×109Nm2/C2)(+1.0μC)(0.60μC(0.060m)3++0.80μC(0.100m)3)(0.060)]i^+[(8.988×109Nm2/C2)(+1.0μC)+0.80μC(0.100m)3(0.080)]j^=[(8.988×109Nm2/C2)(+1.0μC×1C106μC)(0.60μC×1C106μC(0.060m)3++0.80μC×1C106μC(0.100m)3)(0.060)]i^+[(8.988×109Nm2/C2)(+1.0μC×1C106μC)+0.80μC×1C106μC(0.100m)3(0.080)]j^=(1.1N)i^+(0.58N)j^

It is obtained that the magnitude of x and y components of the total force are 1.1N and 0.58N respectively.

Substitute 1.1N for Fx, and 0.58N for Fy in equation (IX) to find Ftotal.

Ftotal=(1.1N)2+(0.58N)2=1.2N

Substitute 0.58N for Fy and 1.1N for Fx in equation (X) to find θ.

θ=tan1(0.58N1.1N)=28°

Therefore, the electric force on the +1.0μC charge due to the other two charges is 1.2N at 28° below the negative x-axis_.

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Chapter 16 Solutions

PHYSICS

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