The dipole moment of a dipole in a 300 N/C electric field is initially perpendicular to the field, but it rotates so it is in the same direction as the field. If an electric dipole moment has a magnitude of 2.0 nC m, the work done by the field is: A. 1.2 nJ В. О C. 0.6 nJ D. -0.6 nJ E. -1.2 nJ

The dipole moment of a dipole in a 300 N/C electric field is initially perpendicular to the field, but it rotates so it is in the same direction as the field. If an electric dipole moment has a magnitude of 2.0 nC m, the work done by the field is: A. 1.2 nJ В. О C. 0.6 nJ D. -0.6 nJ E. -1.2 nJ

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: Rank the electric potential energies in order, from most positive (largest or rank 1) to most…

A: Potential energy of dipole is U = - P E cos(theta)

Q: A 6.0-mF capacitor charged to 50 V and a 4.0-mF capacitor charged to 34 V are connected to each…

A: Two capacitors are charged to some potential difference and then connected in parallel to each…

Q: Two small charged particles have equal charges of + 1.0 nC. They are positioned so that their…

A: The potential charge between two point charges is given by, U=kq1q2r Here, k is the Coulomb…

Q: EXERCISE A constant electric field with magnitude 1.50 x 10³ N/C is pointing in the positive…

A: Concept: The magnitude of a constant electric field is given to be 1.50×103 N/C in the positive…

Q: a. Calculate the maximum torque experienced by this electric pole! b. If the direction of the dipole…

Q: there are 3 charges located at the corners of a rectangle. What is the electric potential energy of…

A: The position of the charges is shown below.

Q: A 3.5πF capacitor is connected to a 12V battery and allowed to fully charge. The capacitor is then…

Q: A capacitor with capacitance C = 6.5 μF has U = 10.5 mJ stored in it during the steady state. A.…

A: Capacitance C=6.5 uFEnergy U=10.5 mJ

Q: A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What…

Q: A parallel-plate capacitor consists of two identical, parallel, conducting plates each with an area…

A: Given data: Plate area (A) = 5.0 cm2 = 5.0×10-4 m2 Charge (q) = 8.4 pC = 8.4×10-12 C Distance (d) =…

Q: The electric potential difference between the ground and a cloud in a particular thunderstorm is 7.7…

A: This problem can be solved using the formula, U=qV where, U is magnitude of potential energy, q is…

Q: Question 13. Two parallel plates are 3.00 cm apart and there is an electric potential difference of…

A: 13) distance between two parallel plates (d) = 3 cm d = 0.03 m potential difference = 44 volts

Q: An ion accelerated through 180.00 V has its potential energy decreased by potential energy of 5.76…

Q: 8. A charge (q = 5 mC) is released from rest in an electric field (vector) of E =2.5* N/(mu*C) i hat…

A: 1) 37.5 J 2) 7.071 m/sExplanation:

Q: An electric field E=10 V/m permeates a region in outer space. What is the electric energy stored in…

Q: What is the magnitude of the electric field potential is given by V = 9.90xyz2, where

Q: Three charges Q1= -8 µC, Q2 = +5.8 µC and Q3 = +2.8 µC are separated between Q1 and Q2 by 8 cm. a…

A: Given: Three charges: Q1=-8μC ; Q2=+5.8μC ; Q3=+2.8μCdistance between Q1 and Q2 d1=8cm To find:…

Q: In the figure an electric dipole swings from an initial orientation i (8₁ = 19.0°) to a final…

Q: n electron is fired in the positive y-direction from x= -3.00 cm in a constant electric field with…

A: Given: X=-3 cm E=1.88×103 N/C X=2 cm

Q: For an electron moving in a direction opposite to the electric field A. its potential energy…

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

The dipole moment of a dipole
in a 300 N/C electric field is
initially perpendicular to the
field,
but it rotates so it is in the same
direction as the field. If an
electric dipole moment has a
magnitude of
2.0 nC m, the work done by the
field is:
A. 1.2 nJ
B. 0
C. 0.6 nJ
D. -0.6 nJ
E. -1.2 nJ

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

A parallel-plate capacitor consists of two identical, parallel, conducting plates each with an area of 4.00 cm2 and uniform charges of ±5.00 nC. The plates are separated by a perpendicular distance of 1.50 mm. What is the potential difference across the metallic plates? a. 3.45 kvolts b. 4.72 kvolts c. 1.08 kvolts d. 2.12 kvolts e. 1.69 kvolts
Suppose had a uniform electric field, E = 720 V/m, that points to the right as shown in the figure. What is the potential difference (multimeter reading) for the following?
The magnitude of the charge moving between the plates is | q | = 10 µC. The voltage between the plates is 15 V, and the electric field between the capacitor plates is 70 N/C. When charge q crossed from the top to the bottom plate how much kinetic energy in Joules does it gain? a. 7.0e-4 J b. 150 J c. 1.5e-4 J d. 42.0 J
A parallel-plate capacitor has a plate separation d320 mm. A potential difference V=1129 volts exists between the plates. What is the magnitude of the force (in Newton) on a proton between the plates?. The charge of proton e=1.6*10-19 C. A. 1.806-10-14 B. 9.504-10-8 C. None D. 4.516-10-15 E. 9.032-10-15
particle of charge = 7uC is located on the -axis at the point = 5.1 cm. A second particle of charge = -4uC is placed on the -axis at = -3.4 cm. Take Coulomb's constant k as. Calculate the net electric potential at the origin. O A. -2240000 O B. None of the other three options OC. 204000 O D. -2640000
A small positive test charge of 0.02C is situated 1.5m away from a positive charge of 1C and 4.2m away from a second positive charge of 14C. What is the total potential energy of the test charge. The coulomb constant K = 9E9. Give your answer in GigaJoules (GJ)
k = Coulomb’s constant
5. A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A 112.0-µC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm). (a) What is the change in the potential energy of the charge-field system?
A uniform electric field with magnitude of 8.00 * 104 that is directed vertically upward has a m charge of 40.0 nC placed inside. What is the change in potential energy of moving the charge a. 0.65 m to the right b. 0.85 m vertically upward C. 0.40 m downward at an angle of 25° from the horizontal
5. You found that the electric potential due to a dipole oriented along the z axis is given by pco$e) 4те a. What is the direction of the field along the x axis? b. If we are on the x axis, a derivative with respect to z is the same as 1 d dz ф Use this equivalence to calculate Ez along the x axis. This should agree with your result from Unit VII p. 20 in your activity guide