VA HW 07 (Freeman X C Chegg.com O I webassign.net/web/Student/Assignment-Respons

Question

VA HW 07 (Freeman X C Chegg.com O I webassign.net/web/Student/Assignment-Responses/submit?dep-13135653#Q25 0.5/2.5 points l Previous Answers My Notes I An electric field given by E 6.0 i- 3(y2 4.0) j pierces the Gaussian cube of the below figure. (E is in newtons per coulomb and X is in meters.) What is the electric flux through the following surfaces? Gaussian surface x 0 m x 3.0 m (a) the top face N m2/C (b) the bottom face N m2/C (c) the left face N m2/c (d) the back face N m (e) the net electric flux through the cube N m2/c For each face, did you use an area vector that is directed outward from the cube? Did you take a dot product of that vector and the electric field vector that erces the area? 11:36 PM O m Cortana. Ask me anything 3/8/2016

Explanation / Answer

Here ,

E = 6 i - 3(y^2 + 4) j N/C

a)

electric flux through top face = E.Area

electric flux through top face = (6 i - 3(2^2 + 4) j).(2^2 j)

electric flux through top face = -96 N.m^2/C

the electric flux through top face is -96 N.m^2/C

b)

electric flux through the bottom face = E .Area

electric flux through the bottom face = - (6 i - 3(0^2 + 4) j).(2^2 j)

electric flux through the bottom face = 12 * 4 N.m^2/C

electric flux through the bottom face is 48 N.m^2/C

c)

electric flux through the left face = - (6 i - 3(2^2 + 4) j).(2^2 i)

electric flux through the left face = -6 * 4 N.m^2/C

electric flux through the left face is -24 N.m^2/C

d)
electric flux through the back face = - (6 i - 3(2^2 + 4) j).(2^2 k)

electric flux through the back face = 0 N.m^2/C

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.