The electric field applied (E) = 7.8 times 104 N/C The vertical rectangular area

Question

The electric field applied (E) = 7.8 times 104 N/C The vertical rectangular area (A) = 0.3 times 0.1m2 The flux through the vertical rectangular area = EAcos180degree = -2.34 times 103 Nm2/C The length of the slanted surface l = 0.10/cos60degree The area of the slanted surface A = (0. times 0.1)/cos60degree = 0.060 m2 The normal to the area is at 60degree with the field direction, the electric flux through the slanted surface = 7.8 times 104 times 0.0.60 times Nm2 / C = 2.34 times 103 Nm2/C The flux through the electric surface of the box (Phi) = flux (rectangular surface +slanted surface + two sides + base) =2.34 times 103+(-2.34 times103) + 0+ 0 +0 Phi =0

Explanation / Answer

Flux = dot product of electric field and area Here electric field and normal to the vertical rectangulararea are anti parallel to each other.So, the angle betweenelectriuc field and area is 180 o.Therefore we put cos 180 o

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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