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Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by \(3.1 \mathrm{mm},\) forming an electric dipole. (a) Find the electric dipole moment (magnitude and dircction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{N} \cdot \mathrm{m} ?\)

Short Answer

Expert verified
(a) Dipole moment is \(1.395 \times 10^{-11} \text{ C} \cdot \text{m}\); (b) Electric field is \(8.61 \times 10^{3} \text{ N/C}\).

Step by step solution

01

Define the Electric Dipole Moment

The electric dipole moment \(\mathbf{p}\) is defined as the product of one of the charges (since they are equal in magnitude) and the distance \(d\) separating them. Given charges \(q_1 = -4.5\, \text{nC}\) and \(q_2 = +4.5\, \text{nC}\) and separation \(d = 3.1\, \text{mm}\), convert \(d\) to meters \(d = 3.1 \times 10^{-3}\, \text{m}\). Then, \( p = q \times d = (4.5 \times 10^{-9} \text{C}) \times (3.1 \times 10^{-3} \text{m}) \).
02

Calculate the Electric Dipole Moment

Substitute the values into the dipole moment formula: \[ p = 4.5 \times 10^{-9} \text{ C} \times 3.1 \times 10^{-3} \text{ m} = 13.95 \times 10^{-12} \text{ C} \cdot \text{m} = 1.395 \times 10^{-11} \text{ C} \cdot \text{m}. \] The direction of the dipole moment is from \(q_1\) to \(q_2\), from negative to positive charge.
03

Relate Torque to Electric Field and Dipole Moment

The torque \(\tau\) on a dipole in a uniform electric field \(\mathbf{E}\) is given by \(\tau = pE \sin \theta\), where \(\theta = 36.9^{\circ}\). Rearrange to find \(E\): \[ E = \frac{\tau}{p \sin \theta}. \]
04

Solve for the Electric Field Magnitude

Substitute \(\tau = 7.2 \times 10^{-9} \text{ N} \cdot \text{m}\), \(p = 1.395 \times 10^{-11} \text{ C} \cdot \text{m}\), and \(\theta = 36.9^{\circ}\) into the equation: \[ E = \frac{7.2 \times 10^{-9}}{1.395 \times 10^{-11} \times \sin(36.9^{\circ})}. \] Calculate \(\sin(36.9^{\circ})\), which is approximately 0.6. Compute the magnitude of \(E\).
05

Compute the Result

Substitute \(\sin(36.9^{\circ}) \approx 0.6\) into the equation: \[ E = \frac{7.2 \times 10^{-9}}{1.395 \times 10^{-11} \times 0.6} \approx 8.61 \times 10^{3} \text{ N/C}. \]
06

Conclusion

(a) The electric dipole moment is \(1.395 \times 10^{-11} \text{ C} \cdot \text{m}\), directed from the negative to the positive charge. (b) The magnitude of the electric field is approximately \(8.61 \times 10^{3} \text{ N/C}\).

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Key Concepts

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

Electric Dipole Moment
An electric dipole consists of two equal but opposite charges, separated by a specific distance. The electric dipole moment \( \mathbf{p} \) quantifies the strength and direction of this dipole. It points from the negative charge to the positive charge which means from \( q_1 \) to \( q_2 \). The formula for calculating the electric dipole moment is:
  • \( p = q \times d \)
where \( q \) is the magnitude of one charge, and \( d \) is the distance between the charges. In the current scenario, this would be: \( p = (4.5 \times 10^{-9} \text{ C}) \times (3.1 \times 10^{-3} \text{ m}) \).
The units for dipole moment are Coulomb-meters (C·m), and this physical quantity is critical in understanding the interaction of dipoles in electric fields. It shows how the separation of charge can create an electric field. A larger dipole moment means the charges have a potentially stronger influence in an electric field.
Uniform Electric Field
A uniform electric field has the same strength and direction at every point in space. This simplifies analysis as the force experienced by charges in such a field is consistent across different positions.
The uniform electric field can interact with an electric dipole, leading to torque on the dipole. A real-life analogy might be turning a compass in a magnetic field, where the field 'wants' to align the compass needle in its direction.
In our problem, this uniform electric field applies a torque to the dipole when it's not aligned with the field. The magnitude of the field itself was to be determined from the known torque, allowing us to gauge how strong this field was at exerting influence on the dipole.
Torque on Dipole
Torque is a measure of the rotational force on an object, in this case, the dipole. When placed in a uniform electric field, a dipole experiences torque because of the forces acting on the two opposite charges.
Torque \( \tau \) is calculated with the formula:
  • \( \tau = pE \sin \theta \)
Here, \( \tau \) is the torque, \( p \) is the electric dipole moment, \( E \) is the electric field strength, and \( \theta \) is the angle between the dipole moment and the electric field. The sine function collaborates the angle's proportion in leading to the rotational effect thus, \( \sin(36.9^{\circ}) \approx 0.6 \).
In our example, the angle wasn’t zero, meaning the dipole wasn’t aligned with the field, resulting in a measurable torque of \( 7.2 \times 10^{-9} \text{ N·m} \).
Electrostatics
Electrostatics studies charges at rest, typically involving electric forces, fields, and potentials caused by stationary charges. It focuses on understanding the forces and reactions between electrically charged objects under various conditions.
This domain helps us understand the behavior of electric dipoles within electric fields. An electric dipole, such as our example, perfectly ties with electrostatic basics since it comprises two static point charges, one positive and one negative.
Analyzing how such dipoles interact with electric fields under the framework of electrostatics can unravel fundamental electromagnetic phenomena that are intrinsically critical in technology, from capacitors to the molecular polarity in chemistry.
Physics Problem Solving
Physics problem solving can be navigated effectively using a systematic approach:
  • Understand the problem: Clarify what’s known and what’s being asked.
  • Use relevant equations: Identify formulas that connect known quantities to unknowns.
  • Substitute values: Plug known values into formulas to find unknowns.
  • Check the units: Ensure your computational steps maintain unit consistency.
  • Verify solutions: Consider if the computational results logically fit the physical scenario.
In our scenario, determining the electric dipole moment and the field required correlating provided values of force and charge to find the electric field strength. By incorporating principles from electrostatics and understanding the interaction of forces, solutions are more intuitive and physics problems become less of a mystery.

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Most popular questions from this chapter

(a) Assuming that only gravity is acting on it, how far does an electron have to be from a proton so that its acceleration is the same as that of a freely falling object at the earth's surface? (b) Suppose the earth were made only of protons but had the same size and mass it presently has. What would be the acceleration of an electron released at the surface? Is it necessary to consider the gravitational attraction as well as the electrical force? Why or why not?

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{cm}\) . (a) Calculate the electric fields \(\vec{E}_{1}\) and \(\overrightarrow{\boldsymbol{E}}_{2}\) at point \(P\) due to the charges \(q_{1}\) and \(q_{2}\) . Express your results in terms of unit vectors (see Example \(21.6 ) .\) (b) Use the results of part (a) to obtain the resultant field at \(P\) , expressed in unit vector form.

A positive point charge \(q\) is placed on the \(+y\) -axis at \(y=a\) and a negative point clarge \(-q\) is placed on the \(-y\) -axis at \(y=-a\) . A negative point charge \(-Q\) is located at some point on the \(+x\) -axis, (a) In a free- body diagram, show the forces that act on the charge \(-Q\) . (b) Find the \(x\) - and \(y\) -components of the net force that the two charges \(q\) and \(-q\) exert on \(-Q .\) (Your answer should involve only \(k, q, Q, a\) and the coordinate \(x\) of the third charge. \()\) (c) What is the net force on the charge \(-Q\) when it is at the origin \((x=0)\) ? (d) Graph the \(y\) -component of the net force on the charge \(-Q\) as a function of \(x\) for values of \(x\) between \(-4 a\) and \(+4 a\) .

Two very large parallel sheets are 5.00 \(\mathrm{cm}\) apart. Sheet \(A\) carries a uniform surface charge density of \(-9.50 \mu \mathrm{C} / \mathrm{m}^{2}\) , and sheet \(B\) , which is to the right of \(A,\) carries a uniform charge of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2} .\) Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 \(\mathrm{cm}\) to the right of sheet \(A\) ; (b) 4.00 \(\mathrm{cm}\) to the left of sheet \(A ;(c) 4.00 \mathrm{cm}\) to the right of sheet \(B\) .

Two small spheres spaced 20.0 \(\mathrm{cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{N} ?\)

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