PH504/Test Answers
UNIVERSITY OF KENT
SCHOOL OF PHYSICAL SCIENCES
2008/09
Answer all questions in the 50 allowed minutes.
1. Consider an inner metal sphere with charge +q of radius a and an outer metal shell with
charge –q and radius b (of negligible thickness). Write down expressions for the electric field
and electric potential in the three regions: (i) within the sphere (ii) between the sphere and the
shell and (iii) external to the shell. Sketch graphs of the radial distributions (please label your
axes). [40 marks]
b
a +q
(i)
(ii)
-q
Inside the sphere, the electric field magnitude E = 0. All charge resides on
the surface since it is a conductor. Therefore V is a constant.
Use the symmetry to consider a Gaussian surface between the shell and
sphere. Charge within the Gaussian surface is +q. Therefore, the electric
flux through the Gaussian surface is +q/o and so …….
E=Q/(40r2) 
V = - Q/(40r) + any arbitrary constant
….as if entire charge were concentrated at the centre.
Return to (i); potential within sphere must be constant and equal to surface potential
= - Q/(40a) + same arbitrary constant
(iii) Apply the Gaussian surface outside the shell. Total charge is zero within. So E=0
outside the shell. Hence V is a constant outside,independent of r, V = - Q/(40b), so
it can match the value at r=b.( Then the arbitrary constant above can be set to zero
also – why not).
Sketches: similar to that in Lecture Notes 5 (except should be extended to beyond
the shell, with E jumping back down to zero again)
2
2. State what the Method of Images can be employed to simplify. Provide sketches to
demonstrate the method. [20 marks]
It simplifies the calculation of the Electric Field between point charges and a plate or
spherical surface, y substituting imaginary charges at appropriate locations. See
Lecture Notes 8 for sketches.
3. State how capacitance C, charge Q and potential V are related. [10 marks]
C = Q/V
4. In a dielectric medium, the differential form of Gauss’s law is
.
Define the two quantities D and f . [10 marks]
D is the electric displacement, or D-field
f is the volume density of free charges.
5. A point charge Q is placed at the centre of a cube in a medium with permittivity . What is
the flux of the E-field over each face? If the charge is moved to the corner of the cube what is
the new flux over each face? Explain your reasoning. [20 marks]
3
The total flux of a point charge is q/ .The cube has 6 sides. Therefore, by symmetry,
the flux through each face is q/(6.
Move the charge, say to (0,0,0)……. There are now 8 possible cubes which would fill
up the entire space around the point charge. We are considering one of the cubes.
(e.g. The cube apices are located at (1,1,1), (1,1,-1), (1,-1,1), (1,-1,-1) and 4 more…
(-1,1,1), (-1,1,-1), (-1,-1,1), (-1,-1,-1). Each of these has 3 sides opposite to the
charge. Therefore the flux through these faces is 1/24 of the total flux: q/(24
The flux through the faces which touch the charge are all zero.
turn over