College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Expert Solution
arrow_forward
Step 1
In this problem expression of electric potential is given as :
V = -5x² + 2xy + 7z
We know,
E = - [(δV/ δx) i + (δV/ δy) j + (δV/ δz) k] ------(1)
Computing the derivative of 'V'
δV /δx = -10x
δV /δy =2x
δV /δz = 7
Putting value of δV in above equation
E = - [ (-10x) i + (2x) j + 7 k ]
E = [(10x) i - (2x) j - 7k ] N/C
where i,j,k are unit vectors in x,y,z directions.
Step by stepSolved in 2 steps
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
- What is the potential difference between xi = 10 cm and xf = 40 cm in the uniform electric field Ex = 2000 V/m?arrow_forwardThe plates of a parallel-plate capacitor are separated by 0.12 mm. The permittivity of a vacuum is 8.85419 × 10-¹2 C²/N. m². If the space between the plates is air, what plate area is required to provide a capacitance of 9 pF? Answer in units of cm².arrow_forwardConsider a parallel-plate capacitor constructed of two square plates, 4.22 m on a side, separated by a distance 6.23 mm. If a charge of 2.85 nC is placed on one plate, with an equal but opposite charge on the other plate, find the electric field inside the capacitor. Answer in units of V/m.arrow_forward
- What is the magnitude of the electric field at the point (2.90 i-7.20 j+5.50 k) m if the electric potential is given by V = 5.30xyz², where Vis in volts and x, y, and z are in meters? Number i Unitsarrow_forwardA charged capacitor consisting of two parallel plates has an area and separate of 0.500 m2 and 9.00 mm respectively. If the electric field strength between the plates is 3.0 x 106 Vm-1, calculate the energy stored in the capacitor.arrow_forwardCathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.56 cm. If the potential difference across the plates was 24.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates. ????V/marrow_forward
- The electric field strength between two parallel conducting plates separated by 3.60 cm is 4.90 ✕ 104 V/m. a) What is the potential difference between the plates (in kV)? and b)The plate with the lowest potential is taken to be at zero volts. What is the potential (in V) 1.00 cm from that plate (and 2.60 cm from the other)?arrow_forwardAn electron moving parallel to the x axis has an initial speed of 3.7 x 106 m/s at the origin. Its speed is reduced to 1.4 x 105 m/s at x = 2 cm. Calculate the electric potential difference 1. between the origin and x -= 2 cm. 2. A proton is released from rest in a uniform electric field whose magnitude is 5000 V/m. Through what potential difference will it have passed after moving 0.25 meters? How fast will it be going after it has travelled 0.25 meters?arrow_forwardThe electric field Strength between two parallel conducting plates separated by 4.60 cm is 8.90*10^4 V/m. a) what is the potential difference in plates (in kV) b) the plate with the lowest potential is taken to be at zero volts. What is the potential (in V) 2cm from that plate and 2.60cm from the other?arrow_forward
- 57. A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has a radius of ra = 12.0 cm , and the outer sphere has a radius of rb = 15.1 cm . A potential difference of 120 V is applied to the capacitor. What is the capacitance of the capacitor? What is the magnitude E1 of the electric field E at radius r= 12.8 cm , just outside the inner sphere? What is the magnitude of E at r= 14.7 cm , just inside the outer sphere?arrow_forwardWhat is the magnitude, in volts, of the maximum potential difference between two parallel conducting plates separated by 0.51 cm of air?arrow_forwardThe voltage across a capacitor is given by the formula V=Q/C, where Q is usually called "the charge on the capacitor." Where is this charge in a capacitor? Does the capacitor really have a net charge?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON