Electric Field in Regions - General Physics - Solved Past Paper, Exams of Physics

Download Electric Field in Regions - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 4. (25 pts) A solid, insulating sphere of radius a has a uniform charge density ρ and total charge Q. Concentric with this sphere is a hollow, uncharged, conducting sphere whose inner and outer radii are b and c, as shown below. a b c Insulator Conductor ra rb rc rd Compute the electric field in the regions (a) r < a, (b) a < r < b, (c) b < r < c, and (d) r > c. You may write your answers in terms or ρ or Q, whichever you prefer. a) r < a This is within the solid sphere of charge. Thus not all of the charge on this sphere will be inside the Gaussian surface of radius ra as shown on the figure in blue. Thus qemc 6= Q. It is most likely easiest to work out the enclosed charge from the charge density ρ, qenc = ρV = 4πr3aρ 3 Putting this into Gauss’s Law gives 4πr2aE = 4πr3 a ρ 3ǫ0 E = ρ ra 3ǫ0 Given the positive charge, the electric field direction is outward. b) a < r < b This is inside the empty space between the solid sphere and the inner surface of the hollow sphere. This time, all the charge on the solid sphere is enclosed by the Gaussian surface of radius rb. This gives qenc = Q. So Gauss’s Law gives 4πr2bE = Q ǫ0 E = Q 4πǫ0 r 2 b c) b < r < c This is inside the conducting shell. Thus E = 0. d) r > c This is outside the entire arrangement at rd on the figure above. The total charge within this Gaussian surface is still Q, giving an electric field E = Q 4πǫ0 r 2 d