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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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Most popular questions from this chapter

Performing an experiment similar to Millikan's oil drop experiment, a student measures these charge magnitudes: \(3.26 \cdot 10^{-19} \mathrm{C} \quad 5.09 \cdot 10^{-19} \mathrm{C} \quad 1.53 \cdot 10^{-19} \mathrm{C}\) \(6.39 \cdot 10^{-19} \mathrm{C} \quad 4.66 \cdot 10^{-19} \mathrm{C}\) Find the charge on the electron using these measurements.

Another unit of charge is the electrostatic unit (esu). It is defined as follows: Two point charges, each of 1 esu and separated by \(1 \mathrm{~cm}\), exert a force of exactly 1 dyne on each other: 1 dyne \(=1 \mathrm{~g} \mathrm{~cm} / \mathrm{s}^{2}=1 \cdot 10^{-5} \mathrm{~N}\). a) Determine the relationship between the esu and the coulomb. b) Determine the relationship between the esu and the elementarv charoe

Rubbing a balloon causes it to become negatively charged. The balloon then tends to cling to the wall of a room. For this to happen, must the wall be positively charged?

A bead with charge \(q_{1}=1.27 \mu \mathrm{C}\) is fixed in place at the end of a wire that makes an angle of \(\theta=51.3^{\circ}\) with the horizontal. A second bead with mass \(m_{2}=3.77 \mathrm{~g}\) and a charge of \(6.79 \mu \mathrm{C}\) slides without friction on the wire. What is the distance \(d\) at which the force of the Earth's gravity on \(m_{2}\) is balanced by the electrostatic force between the two beads? Neglect the gravitational interaction between the two beads.

How far must two electrons be placed on the Earth's surface for there to be an electrostatic force between them equal to the weight of one of the electrons?
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