An electric field given by E = 4.5i -3.6( y 2 + 2.8)·j pierces a Gaussian cubeof

Question

An electric field given by E = 4.5i -3.6(y2 + 2.8)·j pierces a Gaussian cubeof edge length 4.2 m and positioned as shown in the figure below.(The magnitude E is in newtons per coulomb and the positionx is in meters.) What net charge (in Coulombs) is enclosedby the Gaussian cube? Help Please An electric field given by E = 4.5i -3.6(y2 + 2.8)·j pierces a Gaussian cubeof edge length 4.2 m and positioned as shown in the figure below.(The magnitude E is in newtons per coulomb and the positionx is in meters.) What net charge (in Coulombs) is enclosedby the Gaussian cube? Help Please

Explanation / Answer

    we use = integral E . dA     the side length of the cube is 4.2 m     the net enclosed charge is given by     on the top face of the cube y = 4.2 m and     dA = (dA) j     solve for E and for the flux     the charge will b e     q = o        = (8.85 x 10-12C2 / N . m2) ()        = ........ C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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