PM End Date: 3/4/2018 11:59:00 PM (8%) Problem 4: The electric potential in a ce

Question

PM End Date: 3/4/2018 11:59:00 PM (8%) Problem 4: The electric potential in a certain area varies with position as V/m, and c 1 3 V. ax2-bx + c, where a = 2.5 V/m2, b = 11 20% Part (a) Find the electric field vector E in this area in terms of the given variables. E=|( 2 a x-b)I Potential 85% Submissions per attempt) 1|2|3| 0 END I give up! field is the M s/ 20% Part (b) what is the magnitude of the electric field, in volts per meter, at x = 1.0 m? 20% Part (c) what is the direction of the electric field at x = 1.0 m? 20% Part (d) What is the magnitude of the electric field, in volts per meter, at x = 50m? q) 20% Part (e) what is the direction of the electric field at x = 5.0 m? 2

Explanation / Answer

a] Electric field is negative gradient of Potential so in this case, the electric field will be negative derivative of potential function [you seem to have forgotten the negative sign].

Therefore, E = - dV/dx

=> E(x) = - [2ax - b]

at x = 5m,

E = - [2(2.5)(5) - 11] = - 14 V/m

d] The magnitude of Electric field will be:

|E| = 14 V/m.

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