Download Quiz 2 - Rutgers Physics

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