EMF
Exercise Class 4
Answers
1: Long plastic (i.e., insulating) rod. Radius a. Uniform charge density inside.
Charge per unit length = .
Note: The derivation of the electric field is given on the question sheet.
By symmetry, the field lines will radiate outwards, perpendicular to the axis.
E
E
r
a
r
L
AXIAL VIEW
SIDE VIEW
Use Gauss's Law:
1.
Diagram with field pattern: as above.
2.
Best Gaussian surface: Choose a cylinder, radius r (two cases: r < a and r > a).
For either of the two Gaussian cylinders,


Flat ends: E and dA are
perpendicular  E  dA = 0
Curved side: E and dA are parallel
 E  dA = EdA
dA
E
E
E
E
E
dA
dA
and: because r is the same over the whole of the curved side, so is E.
3.
Work out :
 
 E.dA   E.dA +  E.dA +  E.dA  0 + 0 +  E.dA =  EdA
top
bottom
side
So,
 = (E)(Area of curved surface) = (2rL)E
4. Find Qenclosed:
side
side
= E
 dA
side
(i) r < a:
Volume of Gaussian cylinder
Volume of rod
Q enc  (Ch arg e on length L of rod ) x

Q enc
r 2 L
 (L)
a 2 L

r2
L
a2
(i) r > a: Qenc = (Charge on length L of rod) = L
5.
Equate  and Qenc/o to find E:
(i) r <a:
(2rL)E = (r2/a2)L/o 
(ii) r >a: (2rL)E = L/o

E( r ) 
r
2 a 2  0
E( r ) 

2  0 r
r
To find V:
V 

E .d L for any path
0
Never mind the sign, just work out the magnitude of V for the moment.
Choose a radial path from the axis to r (r > a):
dr  dL and E and d L are always parallel.
Break the integral into two parts, 0  a and a  r using the two expressions for E:
V 

a
0

rdr 
2  o a 2
This integrates directly to give V  Vo  Vr 

a
r

dr
2  o r

1  2Log e ( r / a)
4 o
Use common sense to decide on the sigh of V:
V is positive because we would need to push a positive charge from radius r inwards
towards the axis.
2:
V(x,y,z) = 3x2z + 3y2z - 2z3
(a)
 V ˆ V ˆ V ˆ 
E ( x, y, z)   
i
j
k
y
z 
 x
E ( x, y, z)   (6xz )ˆi  (6 yz )ˆj  (3x 2  3y 2  6z 2 )kˆ
(b)
V(0,0,0) = 0
V(1,2,3) = 3(12)3 + 3(22)3 - 2(33) = 9 + 36 - 54 = -9 Volts
The magnitude of the work done in bringing -2C from the origin to (1,2,3) is
|W| = |(Charge)(Potential difference)| = |(-2)(-9)| = |18| Joules
Is W positive or negative:
Positive charge would tend to be pulled by the field from the origin to (1,2,3)
because (1,2,3) has a lower potential.
Negative charge would need to be pushed from the origin to (1,2,3,)
Therefore W is positive:
(c)
W = 18 Joules.
Equipotential line  V(x,yz) is a constant.
Plane parallel to the x-y plane  z is a constant.
V(x,y,z) = 3x2z + 3y2z - 2z3
So 3z(x2 + y2) = V(x,y,z) + 2z2
Therefore
x y
2
2
V(x, y, z)  2z 2

 K , a constant.
3z
This is the equation of a circle with radius K1/2 centred on the origin.