Help Assignment 2 Begin Date: 2/102018 4:00:00 PM--Due Date: 2/26 2018 11:59:00

Question

Help Assignment 2 Begin Date: 2/102018 4:00:00 PM--Due Date: 2/26 2018 11:59:00 PM End Date: 2/28/2018 11:59:00 PM ax2-bx + c, where o = 27 vm2. b = 13.5 (5%) Problem 9: The electric potential in a certain area varies with position as pro V/m, and c =11.5V. 20% Part (a) Find the electric field vector E in this area in terms of the given variables. Grade Summary 0% 100% Potenstial Attempts remining: 3 (2% per attempt) detailed vie SuabenitHit give ap! Hints: 0% deduction per hint. Hints remaining: Feedback: 0% deduction per feedback. 20% Part (b) what is the magnitude of the electric field, in volts per meter, at,-1.0 m? 20% Part (c) what is the direction of the electric field at x = 1.0 m? 20% Pant (d) what is the magnitude of the electric field, in volts per meter, at x-5.0 m? 20%) Part (e) What is the direction of the electric field at x = 5.0 m? -

Explanation / Answer

Given

potential is

V(x) = ax^2-bx+c , a = 2.7 V/m2 , b = 13.5 v/m , c = 11.5 V

Part a

we have the relation between field and potential is  

E = dV/dx

E = (d/dx)(ax^2-bx+c)

E = (2ax-b )

E = (2*2.7*x - 13.5 )

Part b

field at x= 1.0 m is

E(1m) = (2*2.7*1 - 13.5 ) = -8.1 V/m

part C

The direction is , the field is negative means the direction is along -ve x direction

Part d

at x = 5 m is  

E(5 m) = (2*2.7*5 - 13.5 ) = 13.5 V/m

Part e

The direction is , the field is positive means the direction is along +ve x direction

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