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An electric dipole has opposite charges of \(5.00 \cdot 10^{-15}\) C separated by a distance of \(0.400 \mathrm{~mm} .\) It is oriented at \(60.0^{\circ}\) with respect to a uniform electric field of magnitude \(2.00 \cdot 10^{3} \mathrm{~N} / \mathrm{C}\). Determine the magnitude of the torque exerted on the dipole by the electric field.

Short Answer

Expert verified
Question: Calculate the magnitude of the torque exerted on an electric dipole with charges q = \(5.00 * 10^{-15}\) C, distance between charges d = 0.400 mm, an electric field magnitude E = \(2.00 * 10^{3} \, N/C\), and angle θ = \(60.0^{\circ}\). Answer: The magnitude of the torque exerted on the electric dipole by the electric field is \(1.73 * 10^{-15} \, N \cdot m\).

Step by step solution

01

Find the electric dipole moment

The electric dipole moment (p) can be calculated using the formula: \(p = qd\) where q is the charge and d is the distance between the charges. We are given q = \(5.00 * 10^{-15}\) C and d = 0.400 mm. First, convert the distance from millimeters to meters: \(d = 0.400 \, mm = 0.400 * 10^{-3} \, m\) Now calculate the electric dipole moment: \(p = (5.00 * 10^{-15} \, C) * (0.400 * 10^{-3} \, m) = 2.00 * 10^{-18} \, C \cdot m\)
02

Calculate the torque using the given electric field and angle

Using the electric dipole moment calculated in step 1, the electric field magnitude E = \(2.00 * 10^{3} \, N/C\), and the angle θ = \(60.0^{\circ}\), we can calculate the torque τ using the formula: \(τ = pE \sin θ\) We first convert the angle from degrees to radians: \(θ = 60.0^{\circ} = \frac{60}{180} * π \, rad = \frac{π}{3} \, rad\) Now calculate the torque: \(τ = (2.00 * 10^{-18} \, C \cdot m) * (2.00 * 10^{3} \, N/C) * \sin(\frac{π}{3}) = 2.00 * 10^{-18} * 2.00 * 10^3 * \frac{\sqrt{3}}{2} = 1.73 * 10^{-15} \, N \cdot m\) Thus, the magnitude of the torque exerted on the electric dipole by the electric field is \(1.73 * 10^{-15} \, N \cdot m\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Dipole Moment
The electric dipole moment provides a quantitative measure of the separation of charge within a molecule or system. It's a vector quantity that points from the negative charge towards the positive charge and its magnitude is given by the product of the charge magnitude (\( q \)) and the distance (\( d \text{ distance between the charges} \text{.}To picture this, consider two equal and opposite charges, like the negative and positive ends of a battery, a little distance apart. The electric dipole moment is how we represent the 'twist' these charges can provide when placed in an electric field.In our solved exercise, the electric dipole moment calculations begin with understanding that the charge and the separation distance are the critical elements. The charges are given as \text{ distance between the charges} \text{ separated by a }\text{ distance between the charges \textwidth, which, when multiplied together, provide the electric dipole moment for the system. Therefore, the electric dipole moment here would be \( 2.00 \times 10^{-18} \, C \cdot m \).In a uniform electric field, the dipole experiences a couple, or pair of opposing forces, that attempt to rotate it which leads us to the calculation of torque.
Uniform Electric Field
A uniform electric field is characterized by equal field strength and direction at every point within it. Think of it as a steadflowting breeze steadily blowing across an entire field, exerting the same influence on every leaf and blade of grass.Using uniform fields simplifies the understanding and calculation of forces on electric charges. We use this concept in our exercise, with a uniform field strength of \(2.00 \times 10^{3} \, \text{N/C} \). It's the consistency of this field that allows us to predict the behavior of an electric dipole when placed within it.In the context of torque exerted on an electric dipole, the uniform electric field provides a steady context to anticipate how much and in what direction the dipole will want to rotate, assuming no other external forces are at play. It ensures that the torque is the same at any point within the field, a crucial simplification for calculations.
Torque Calculation
Torque is a concept originating from mechanics, referring to the rotational equivalent of linear force. In the case of an electric dipole in an electric field, torque describes how strongly the dipole is ‘twisted’ due to the field.The torque on an electric dipole can be calculated with the formula:\[ \tau = pE \sin \theta \]where \( \tau \) is the torque, \( p \) is the electric dipole moment, \( E \) is the electric field strength, and \( \theta \) is the angle between the dipole moment vector and the electric field vector.As demonstrated in the solution, the angle is an important factor as it affects the strength of the torque. At 0 degrees or 180 degrees (when the dipole is perfectly aligned with the field), the sine function would make the torque zero, meaning no rotational force. Yet at 90 degrees, the dipole experiences maximum torque. In the provided exercise, with an angle of 60 degrees, sine of the angle results in a torque calculation that reflects a strong but not maximum turning effect on the dipole.
Electric Charge
Electric charge is the fundamental property of matter that causes it to experience a force when placed in an electric field. There are two types of electric charges, commonly referred to as positive and negative. Like charges repel each other, while opposite charges attract.In the exercise, we're given charges of \(5.00 \times 10^{-15}\) C, which is incredibly small – much smaller than the charge of a single electron. Yet, even such minute charges can create an electric dipole moment when separated by a distance, resulting in significant physical phenomena like torque when placed in an external electric field.By understanding the nature of electric charge, its interaction with electric fields, and how these interactions manifest in physical systems like dipoles, students can better conceptualize fundamental principles of electromagnetism and apply them to a variety of electrical and magnetic situations.

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Most popular questions from this chapter

The six faces of a cubical box each measure \(20.0 \mathrm{~cm}\) by \(20.0 \mathrm{~cm},\) and the faces are numbered such that faces 1 and 6 are opposite to each other, as are faces 2 and \(5,\) and faces 3 and \(4 .\) The flux through each face is given in the table. Find the net charge inside the cube. $$ \begin{array}{|c|c|} \hline \text { Face } & \text { Flux }\left(\mathrm{N} \mathrm{m}^{2} / \mathrm{C}\right) \\ \hline 1 & -70.0 \\ \hline 2 & -300.0 \\ \hline 3 & -300.0 \\ \hline 4 & +300.0 \\ \hline 5 & -400.0 \\ \hline 6 & -500.0 \\ \hline \end{array} $$

A solid metal sphere of radius \(8.00 \mathrm{~cm},\) with a total charge of \(10.0 \mu \mathrm{C}\) is surrounded by a metallic shell with a radius of \(15.0 \mathrm{~cm}\) carrying \(\mathrm{a}-5.00-\mu \mathrm{C}\) charge. The sphere and the shell are both inside a larger metallic shell of inner radius \(20.0 \mathrm{~cm}\) and outer radius \(24.0 \mathrm{~cm} .\) The sphere and the two shells are concentric. a) What is the charge on the inner wall of the larger shell? b) If the electric field outside the larger shell is zero, what is the charge on the outer wall of the shell?

A solid conducting sphere of radius \(r_{1}\) has a total charge of \(+3 Q\). It is placed inside (and concentric with) a conducting spherical shell of inner radius \(r_{2}\) and outer radius \(r_{3}\). Find the electric field in these regions: \(rr_{3}\)

Three charges are on the \(y\) -axis. Two of the charges, each \(-q\), are located \(y=\pm d\), and the third charge, \(+2 q\), is located at \(y=0\). Derive an expression for the electric field at a point \(P\) on the \(x\) -axis.

Two infinite sheets of charge are separated by \(10.0 \mathrm{~cm}\) as shown in the figure. Sheet 1 has a surface charge distribution of \(\sigma_{1}=3.00 \mu \mathrm{C} / \mathrm{m}^{2}\) and sheet 2 has a surface charge distribution of \(\sigma_{2}=-5.00 \mu \mathrm{C} / \mathrm{m}^{2},\) Find the total electric field (magnitude and direction) at each of the following locations: a) at point \(P, 6.00 \mathrm{~cm}\) to the left of sheet 1 b) at point \(P^{\prime}, 6.00 \mathrm{~cm}\) to the right of sheet 1

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