A uniformly charged conducting plate with area A has a total charge Q which is p

Question

A uniformly charged conducting plate with area A has a total charge Q which is positive. The figure below shows a cross sectional view of the plane and the electric field lines due to the charge on the plane. The figure is not drawn to scale. Find the magnitude of the field at point P. which is a distance a from the plate. Assume that a is very small when compared to the dimensions of the plate, such that edge effects can be ignored. ||E_P|| = 4 pi _0 a^2 Q ||E_P|| = Q/4 pi _0 a^2 ||E_P|| = 2 _0 Q A ||E_P|| 4 pi _0 a Q ||E_P|| = Q/2 _0 A ||E_P|| = _0 Q a^2 ||E_P|| = _0 Q A ||E_P|| = Q/_0 A Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q while plate #2 has a charge -3Q. Using the superposition principle find the magnitude of the electric field at a point P in the gasp. ||E_P|| = 4Q/_0 A ||E_P|| = Q/2 _0 A ||E_P|| = 5 Q/_0 A ||E_P|| = Q/_0 A ||E_P|| = 3 Q/2 _0 A ||E_P|| = Q/3 _0 A ||E_P|| = 0 ||E_P|| = 2Q/_0 A ||E_P|| = Q/_0 ||E_P|| = 3Q/_0 A

Explanation / Answer

for point near the large sheet:

E = sigma / 2e0

E = ( Q / A) / 2e0

E = Q / (2 e0 A )

Ans(5)

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

due to 1:

E1 = Q / (2 e0 A) to the right

due to 2:

E2 = 3Q / (2 e0 A) to the right

E = E1 + E2 = 4 Q / (2 e0 A ) = 2 Q / e0 A

Ans(8)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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