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Chapter 22: Problem 41

A 5.0 -g object carries \(3.8 \mu \mathrm{C}\). It acquires speed \(v\) when accelerated from rest through a potential difference \(V\). If a 2.0 -g object acquires twice the speed under the same circumstances, what's its charge?

Short Answer

Expert verified
The charge of the second object is \(1.52 \times 10^{-6} C\).

Step by step solution

01

Understand the problem and identify key information

Two objects are accelerated through the same potential difference but acquire different speeds. The first object's charge and mass are given, and it's necessary to find the charge of the second object. This is a principle of physics: the electrical potential energy provided by the potential difference is transformed into kinetic energy. Here, the following key information is given: Mass of the first object \(m_1 = 5.0g = 0.005kg\), Charge of the first object \(q_1 = 3.8 \mu C = 3.8 \times 10^{-6}C\), Mass of the second object \(m_2 = 2.0g = 0.002kg\), and speed of the second object is twice the speed of the first object \(v_2 = 2v_1\)
02

Apply the relationship between potential and kinetic energy

The energy provided by the potential difference is transformed into kinetic energy of the object. Therefore, the change of potential energy equals kinetic energy, as given by the formula: \(qV = \frac{1}{2}mv^2\). Rearrage this formula to solve for \(V\), we obtain \(V = \frac{1}{2q}mv^2\). Substituting the known values for the first object, \(V = \frac{1}{2 * (3.8 \times 10^{-6}C)} * (0.005kg) v^2 = \frac{0.005kg}{7.6 \times 10^{-6}C}(v^2)\)
03

Use the velocity relation to find the charge of the second object

Since the potential difference is the same for both objects, the potential of the second object equals that of the first object. Using the formula for the potential and expressing \(v_2 = 2v_1\), we have \(\frac{0.005kg}{7.6 \times 10^{-6}C}(v^2) = \frac{m_2}{2q_2}(4v^2)\). Substitute the known values to find \(q_2\), we obtain \(q_2 = \frac{m_2}{2} \times \frac{7.6 \times 10^{-6}C}{0.005kg} = 1.52 \times 10^{-6}C\)

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Most popular questions from this chapter

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