Hollow Spheres Part A Q, Consider the following arrangement of two conducting ho

Question

Hollow Spheres Part A Q, Consider the following arrangement of two conducting hollow spheres with a point charge of Qo 3.30 C at the center. The inner sphere has a radius of 0.011 m and carries a net charge of Q1--1.50 C. The outer sphere has a radius of 0.061m and carries a net charge of Q2-6.70 pC. Calculate the magnitude of the electric field at point A located at a distance 0.021 m from the center Submit Answer Tries 0/15 Part B Calculate the surface charge density on the outer surface of the outer sphere. Neglect the thickness of the sphere. Submit Answer Tries 0f15

Explanation / Answer

A. given two conducting hollow spheres with a point charge Qo at the center

Qo = 3.3micro C

inner sphere, Q1 = -1.5 micro C, r1 = 0.011 m

outer sphere, Q2 = 6.7 micro C, r2 = 0.061 m

electric field at a point at r = 0.0221 m

this point is between the oouter and the inner spheres

consider a spherical gaussean surface at this radius

then from gauss law

E*4*pi*r^2 = qen/epsilon ( where epsilon is permittivity of free space)

here qen is charge enclosed by the gaussean surface, qen = Qo + Q1 = (3.3 - 1.5) micro C = 1.8 micro C

so E = 1.8*10^-6/4*pi*epsilon*r^2 = 8.98*10^9(1.8*10^-6)/0.0221^2

E = 33095145.472 V/m

B. on the outer surface of the outer sphere, let the charge density be sigma

consider a gaussean surface ar radius r2, this gaussean surface lies inside a conductor so electric field has to be 0

so enclosed charge in this gaussean surface is 0

let q be the charge on the inner surface of the outer shell

then Qo + Q1 + q = 0

q = -1.8 micro C

so charge on outer surface of outer shell = Q2 - q = (6.7 + 1.8)micro C = 8.5 micro C

so sigma = 8.5*10^-6/4*pi*0.061^2 = 1.81781*10^-4 c/M^2

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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