(a) Calculate the flux through the cubic closed surface shown in the diagram. Th

Question

(a) Calculate the flux through the cubic closed surface shown in the diagram. The electric field is given by the equation. Units for x and yare in metres. Hint: The problem is similar to the one worked in class, except that the cube has changed dimensions, and the field is a little more complicated. Break the problem into 6 integrals, one for each face of the cube. Two of these integrals are zero. Explain which are zero, and why they are zero. (b) Use Gauss? Law to calculate the charge enclosed by the cube

Explanation / Answer

Cube means all sides are equal

Area of each side is 1*1.5 = 1.5m^2

then ELcric flux is phi_E1= 2*integral of (3x) from x= 1 to x= 1.5
2*(3/2)*x^2 from x= 1to1.5

phi_E1 = 2*1.5*(1.5^2-1^2) = 3.75

phi_E2 = 2*(-y^2/2) from y = 1 to 1.5

phi_E2 = 1.5^2-1 = 1.25

net flux is 1.25+3.75 = 5

------------------------------

by gauss law

Flux = q/eo

charge q = flux*eo = 5*8.85*10^-12 = 44.25 pC

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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