[1 mC = 10–3 C, 1 ?C = 10–6 C, 1 nC = 10–9 C.]Particles of charge +75, +48, and –85 ?C | StudySoup

Textbook Solutions for Physics: Principles with Applications

Chapter 16 Problem 12P

Question

(II) Particles of charge , and  are placed in a line (Fig. 16-49). The center one is  from each of the others. Calculate the net force on each charge due to the other two.

Solution

Solution 12P

Step 1 of 4:

The three charges are kept in a straight line. The distances between the center charge from the first and third charges are equal. Our aim here is to find the net force on each charge due to the other charges.

The first charge Q1 = +75 μC = 75 x 10-6 C

The second charge Q2 = +48 μC = 48 x 10-6 C

The third charge Q3 = -85 μC = -85 x 10-6 C

The distance between charge 1 and 2 (r12) = 0.35 m

The distance between charge 2 and 3 (r23) = 0.35 m

The distance between charge 1 and 3 (r13) = 0.35 m + 0.35 m = 0.70 m

The constant k = 9.0 x 109 N.m2/C2

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full solution

Title Physics: Principles with Applications 6 
Author Douglas C. Giancoli
ISBN 9780321569837

[1 mC = 10–3 C, 1 ?C = 10–6 C, 1 nC = 10–9 C.]Particles of charge +75, +48, and –85 ?C

Chapter 16 textbook questions

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