A solid metal sphere of radius a = 1.30 cm is surrounded by a concentric spheric
ID: 1983713 • Letter: A
Question
A solid metal sphere of radius a = 1.30 cm is surrounded by a concentric spherical metal shell of inner radius b = 2.60 cm and outer radius c = 3.10 cm. The inner sphere has a net charge of Q1 = 3.40 C, and the outer spherical shell has a net charge of Q2 = -8.70 C.
What is Er at a point located at radius r = 3.50 cm, i.e. outside the outer shell?
What is the surface charge density, b, on the inner surface of the outer spherical conductor?
What is the surface charge density, c, on the outer surface of the outer spherical conductor?
A solid metal sphere of radius a = 1.30 cm is surrounded by a concentric spherical metal shell of inner radius b = 2.60 cm and outer radius c = 3.10 cm. The inner sphere has a net charge of Q1 = 3.40 mu C, and the outer spherical shell has a net charge of Q2 = -8.70 mu C. What is the radial component of the electric field Er at a point located at radius r = 2.30 cm, i.e. between the two conductors? Er is positive if it points outward, negative if it points inward. What is Er at a point located at radius r = 3.50 cm, i.e. outside the outer shell? What is the surface charge density, ?b, on the inner surface of the outer spherical conductor? What is the surface charge density, ?c, on the outer surface of the outer spherical conductor?Explanation / Answer
Charge on inner surface of outer conductive shell must cancel charge on central solid conductive sphere in order for E Field to equal zero (0) inside outer conductive shell: {Charge On Inner Surface Of Outer Conductive Shell} = Q = = -{Charge On Central (Inner) SolidConductive Sphere} ----> Q = -(3.7e-3 C) {Surface Charge Density On Inner SurfaceOf Outer Conductive Shell} = = Q/(4*Pi*R^2) = (-3.7e-3 C)/(4*Pi*(2.1e-2m)^2) = -0.6677 C/m^2
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.