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Solutions to Quiz 2 March 22, 2010 1 Ball of Charge A spherical ball of radius R has uniform charge density spread throughout its volume. 1.1 Outside the Ball At what radial distance r outside the ball is the electric field 1/4 of its maximum value? Express your answer in terms of R. Gauss’ Law states that the flux is determined by the enclosed charge: ~ = qenclosed ~ · da E (1) 0 Consider a sphere of radius r > R that surrounds the ball and has the same center. The electric field is constant over this sphere and points along R ~ = EA. The enclosed charge is the charge of ~ · da the area vector, so that E the ball Q. So EA = Q/0 or the electric field is given by Z E= 1 Q 4π0 r2 (2) So if you’re outside the ball of charge, Gauss’ Law tells us the electric field is the same as that of a point charge at its center. This electric field is a maximum when r = R, and is 1/4 of this value when r = 2R (as it falls of with the square of radial distance). 1 2 Inside the Ball At what radial distance r inside the ball is the electric field 1/4 of its maximum value? Express your answer in terms of R. Show your derivation. Now instead consider a sphere of radius r < R that is concentric with the ball. By symmetry and the fact that charge is positive, the electric field must point radially outwards (any other direction would not be spherically R ~ = EA. ~ · da symmetric). So we again have that E So using Gauss Law once again, we get E= 1 Qenclosed 4π0 r2 (3) where Qenclosed is the charge contained inside this sphere. To work out the enclosed charge, we use that Q ρ= 4 πR3 3 i.e. = Qenclosed 4 πr3 3 r = R Qenclosed (4) 3 Q (5) Substituting (5) in (3) gives E= 1 Q r 4π0 R3 (6) So inside the ball, the 1/r2 dependence of electric field and r3 dependence of charge partially cancel, leaving us with E that is increasing linearly in r. It is maximum at r = R. So it is a quarter of the max at r = R4 . 2.1 Plot of E Plot the electric field E as a function of r in the range 0 < r < 4R. 2