It\'s a question related to physics2 Can you please show all the steps Thanks, C

Question

It's a question related to physics2
Can you please show all the steps
Thanks, Chapter 23, Problem 002 Your answer is partially correct. Try again. An electric field given by E 1.4 i 2.0(ya 6.5) j pierces the Gaussian cube of edge length 0.440 m and positioned as shown in the figure. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? Gaussian surface (a) Number 59 Units (b) Number 2.52 Units 2/C (c) Nu -0.27 (d) Number 0 Units This answer has no units (e) Number -0.07

Explanation / Answer

a) flux through the top face = E.A

= (1.4 i - 2*(y^2 + 6.5)j).(0.44^2)j

= -2*(y^2 + 6.5)*0.44^2

= -2*(0.44^2 + 6.5)*0.44^2

= -2.59 N.m^2/C

b) flux through the bottom face = E.A

= (1.4 i - 2*(y^2 + 6.5)j).(-0.44^2)j

= 2*(y^2 + 6.5)*0.44^2

= -2*(0^2 + 6.5)*0.44^2

= -2.52 N.m^2/C

c) flux through left face = E.A

= (1.4 i - 2*(y^2 + 6.5)j).(0.44^2)(-i)

= -1.4*0.44^2

= -0.271 N.m^2/C

d) flux through right face = E.A

= (1.4 i - 2*(y^2 + 6.5)j).(0.44^2)(i)

= 1.4*0.44^2

= 0.271 N.m^2/C

e) net elctric flux through the cube = -2.59 + 2.52 - 0.271 + 0.271

= -0.07 N.m^2/C

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