In a certain region of space, the electric potential isV(1, y, z) = Bzy – Cy² + Azwhere A, B and C are positive constants.Calculate the r, y and z components of the electric field.E, = -Cy+ 2Ar, Ey= -Cx – B , E,=0%3DE = -3Cz + 2 Ar , E, = 0, E, = -3Cx-B%3D%3DE, = -Az + 2Br ,E, = 0 ,E, = - Ax – C

Given Data: The electric potential is, Vx,y,z=Bzy-Cy2+Az. The electric field can be calculated as…

Answered: In a certain region of space, the…

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)...

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a uniform electric is indicated by electric line. the electric magnitude is 150 n/c. if the object is moved along the pathway from point a to point b and the length of the path is 25 centimeters , calculate the charge in electrostatic potential when you move the object along the path.
Two concentric spherical conducting shells are separated by vacuum. The inner shell has total charge +Q and radius a, and outer shell has charge -Q and radius b. Using integration of the electric field energy density find the electric energy stored in the system. Take Q = 28 µC, a = 20 cm and b = 70 cm. I 1 The energy, U₁ = 12.582440✔✔ Units J The energy, U₁ a = = Using the obtained energy and formula for the energy stored in a capacitor, U capacitance of the system. The capacitance, Co 3.1154528 X Units Select an answer ✓ Repeat the calculations for the same system with a dielectric material of k = 4.5 inserted in between the shells. The capacitance, C₁ = -Q Units Select an answer ✓ .37489732 X Units pF Q² 20¹ find the
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A From Equation 25.6, we have the following. AV = -Ed = -(7.2 × 104 v/m)(0.60 m) = B V From Equation 25.3, we have the following. AU = qAV = eAV = (1.6 x 10-19 C)(AV) | The negative sign means that the potential energy of the system decreases as the proton moves in the direction of the electric field. This decrease is consistent with the conservation of energy in an isolated system: as the proton accelerates in the direction of the field, it gains kinetic energy and at the same time the system loses electric potential energy. The increase in kinetic energy of a charged particle in an electric field is exploited in many devices, including electron guns for TV picture tubes and particle accelerators for research in particle physics. You can predict and observe the speed of the proton as it arrives at the negative plate for random values of the electric field using this Interactive Example. Hints: Getting Started | I'm Stuck Exercise 25.2 For the previous example, apply the principle of…
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Problem 1: Consider a parallel-plate capacitor made up of two conducting plates with dimensions 39 mm × 32 mm. Part (c) What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor? E=5.34 * 104 E = 5.340 x 104 ✓ Correct! E' = Part (d) If the separation between the plates doubles, what will the electric field be if the charge is kept constant? tan() I ( ) 7 8 9 HOME acos() EMAL 4 5 6 sinh() 1 2 3 cotanh() 0 NO BACKSPACE sin() cos) cotan() asin() atan() acotan() cosh() tanh() Degrees O Radians + - END DEL CLEAR .
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What is the capacitance of a parallel plate capacitor in vacuum? Sol. The basic definition for capacitance is: C = Since the electric potential can also be written as V = Ed Where, E is the electric field contribution of the parallel plate (two plane charge): = ol E0 %3D Sigma is the surface charge density given by: O= By, substitution, we get the capacitance of the parallel plate to be:
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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 Units
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