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Chapter 21: Problem 6

Which of the following situations produces the largest net force on the charge \(Q\) ? a) Charge \(Q=1 \mathrm{C}\) is \(1 \mathrm{~m}\) from a charge of \(-2 \mathrm{C}\). b) Charge \(Q=1 \mathrm{C}\) is \(0.5 \mathrm{~m}\) from a charge of \(-1 \mathrm{C}\) c) Charge \(Q=1 \mathrm{C}\) is halfway between a charge of \(-1 \mathrm{C}\) and a charge of \(1 \mathrm{C}\) that are \(2 \mathrm{~m}\) apart. d) Charge \(Q=1 \mathrm{C}\) is halfway between two charges of \(-2 \mathrm{C}\) that are \(2 \mathrm{~m}\) apart. e) Charge \(Q=1 \mathrm{C}\) is a distance of \(2 \mathrm{~m}\) from a charge of \(-4 \mathrm{C}\).

Short Answer

Expert verified
Answer: The largest net force is in situation b) with a magnitude of 35.96 x 10^9 N.

Step by step solution

01

Compute force for situation a)

For the situation a), Q = 1C, 1m from a charge of -2C. We have: q1 = 1 C q2 = -2 C r = 1 m Using the Coulomb's Law formula, we get the force Fa: \(Fa = k \frac{q1 \cdot q2}{r^2} = 8.99 \times 10^9\frac{1 \times (-2)}{1^2} = -17.98 \times 10^9 N\)
02

Compute force for situation b)

For the situation b), Q = 1C, 0.5m from a charge of -1C. We have: q1 = 1 C q2 = -1 C r = 0.5 m Using the Coulomb's Law formula, we get the force Fb: \(Fb = k \frac{q1 \cdot q2}{r^2} = 8.99 \times 10^9\frac{1 \times (-1)}{0.5^2} = -35.96 \times 10^9 N\)
03

Compute vector forces for situation c)

For the situation c), Q = 1C is halfway between a charge of -1C and a charge of 1C that are 2 m apart. First, find the distance to each charge: 1 m. Next, find the force caused by the -1C charge and the force caused by the +1C charge. Due to symmetry, these forces will have opposite directions and will cancel each other out. Thus, the net force is 0 N for this situation.
04

Compute vector forces for situation d)

For the situation d), Q = 1C is halfway between two charges of -2C that are 2 m apart. First, find the distance to each charge: 1 m. Calculate the force caused by each -2C charge: \(Fd1 = k \frac{1 \cdot (-2)}{1^2} = -17.98 \times 10^9 N\) Because of symmetry, the two forces have the same magnitude but act in opposite directions. The net force Fd: \(Fd = 2 \times |-17.98 \times 10^9 N| = 35.96 \times 10^9 N\)
05

Compute force for situation e)

For the situation e), Q = 1C, 2m from a charge of -4C. We have: q1 = 1 C q2 = -4 C r = 2 m Using the Coulomb's Law formula, the force Fe: \(Fe = k \frac{q1 \cdot q2}{r^2} = 8.99 \times 10^9\frac{1 \times (-4)}{2^2} = -17.98 \times 10^9 N\) Now we have all net forces for each situation: Fa = -17.98 x 10^9 N, Fb = -35.96 x 10^9 N, Fc = 0 N, Fd = 35.96 x 10^9 N, Fe = -17.98 x 10^9 N. The largest net force will have the greatest magnitude, which is Fb in situation b) with a magnitude of 35.96 x 10^9 N.

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