ZAMBIA UNIVERSITY COLLEGE OF TECHNOLOGY
EEE4020 – ELECTROMAGNETIC FIELD
ELECTROSTATICS – TUTORIAL 1
QUESTION ONE
Consider a long cylindrical charge distribution of radius, r, with a uniform charge density, ρ.
determine, the electric field at a point P, situated at a distance, R, from the center axis. For a
cylinder of radius, r = 7cm, find the magnitude of the field at R = 10cm measured from the center
axis with a uniform charge density of 6.5 X 10-5 C/m3.
Figure 1.1: Cylindrical charge distribution
Solution






The field will have a positive divergence and will radiate outwards in the 𝜌̂ unit vector
direction.
Cylinder is infinitely long and stretches fro – infinity to + infinity.
All flus projects radially
All the charge is uniformly distributed on the cylindrical curved surface.
The bottom and the top parts will have NO flux through them, i.e no flux in the 𝑧̂ unit vector
direction.
The Gaussian surface will be as depicted in Figure 1.2 below
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Figure 1.2: Gaussian Surface
i.
Applying Gauss’s law,
 EnˆdS 
s
Q
0
Where,
𝐸⃗ = Electric field at point P on the Gaussian surface
dS = differential Gaussian surface area
𝑛̂ = unit normal vector in the 𝜌̂ unit vector direction
Q = charge enclosed by the cylinder with radius, r.
έ = permittivity of free space, 8.85 x 10-12
ii.
This equation can be expressed as;
E  dA 
s
Q
0
……………………………(ii)
Where,
dA = 2πRL
E ( 2RL ) 
iii.
Q
0
……………………….(iii)
Solving for electric filed strength, E
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E
iv.
Q
……………………………(iv)
2RL  0
But volume charge density, ρ, is given as;
Q
; Q  V
V
Where,
V = volume of a cylinder, πr2L in m3

Q  r 2 L ……………………..(v)
v.
Replacing eqt. (v) into eqt. (iv) give;
E
r 2 L
2RL  0
r 2
E
2 R 0
vi.
When the Gaussian surface is the same as the cylindrical surface, i.e. r = R
E
vii.
………(vi)
r
2 0
………..(vii)
Therefore, electric field at 10cm from the center axis of a cylinder of radius 7cm will be;
E
(6.5  10 5 )(0.07) 2
r 2

 1.8  105 N / C
12
2 R 0 2(0.1)(8.85  10 )
QUESTION TWO
 109 
C / m 2 are located at z = -5m and y = Two infinite sheets of uniform charge densities  S  
 6 
5m. Determine the uniform line charge density  L necessary to produce the same value of
electrostatic force E at (4, 2, 2)m if the line charge is at y= 0, z = 0.
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QUESTION THREE
Two small identical conducting spheres have charges of 2 nC and -1 nC respectively. When they
are separated by a distance of 4cm. Find the magnitude of the force between them. If they are
brought into contact and the then separated by 4cm, find the force between them.
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QUESTION THREE
a) State divergence theorem


b) A particular vector field H  r 2 cos2 r  z sin  is in cylindrical system. Find the flux
emanating due to this field from the closed surface of the cylinder 0  z  1, r  4 . Verify
the divergence theorem. (BAKSHI p76).
Solution
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QUESTION FOUR
a) State stokes theorem


b) Verify Stoke’s theorem for a vector field A  r 2 cosr  z sinz around the path L defined
by 0  r  3 , 0    45 and z  0 as shown in the Figure below. (BAKSHI, p74)
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Solution
b)
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QUESTION FIVE
A non-conducting sphere of radius R = 7cm carries a charge density of ρ= 10 -9C/m3, distributed
uniformly throughout its volume. At what distance within the sphere, measured from the center of
the sphere does the electric filed reach a value of E =1.32N/m.
Solution
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Applying Gauss’s law,
 EnˆdS 
s
Q
0
 (4 / 3)r 3
E (4R ) 
0
2
But the field is uniformly distributed, hence; R = r
E
1.32 
 (4 / 3)R 3
4R 2 0
E
R
3 0
(10 9 ) R
3(8.85  10 12 )
R  0.035m  3.5cm
QUESTION SIX
a) Given a cylindrical surface, infinite in length with radius r and a uniform charge distribution
 S C/m at its surface. Determine the expression for the electric field at a distance R away
from the center of the cylinder.
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b) An infinitely long cylinder with inner radius 8cm and outer radius 10cm has uniform
volume charge density, ρ = 8.85x10-9 C/m3. Determine the magnitude of the electric field at
a point P, distance of 4cm from the inner radius.
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QUESTION EIGHT
a) A pin type insulator used to isolate a conductor from its physical support on a ZESCO steel
pole has a relative permittivity of 7.2 and thickness 40.2 mm. The conductor is at a potential
of 11000V, and the pole is at ground potential. Due to aging a crack with an air pocket of
thickness 0.25mm develops in the insulator. Clearly defining any symbol used, derive an
expression for the electric field strength in the air pocket as a function of the applied
voltage.
b) Use the formula derived in (a) to determine whether there will be a flash-over in the air
pocket if the dielectric strength of the air is 3kV/mm.
c) A charged paint is spread in a very thin uniform layer over the surface of a soccer ball of
diameter 300mm, giving it a charge of -21nC. Using gauss’s law of electrostatics,
determine;
i.
ii.
iii.
the magnitude of the electric field just inside the paint layer
the magnitude of the electric filed just outside the paint layer
the electric field 90mm outside the surface of the paint layer
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QUESTION NINE
A non-conducting sphere of radius R = 7cm carries a charge of Q= 5 x 10-3C, distributed uniformly
throughout its volume. At what distance within the sphere, measured from the center of the sphere
does the electric filed reach a value equal to half the maximum value.
Solution
Figure 11.1
i.
Applying Gaussian law;
 EnˆdS 
s
E  dA 
s
0
Q
0
The maximum filed strength is;
E max (4R 2 ) 
E max 
iii.
Q
0
Q
5  10 3

 9.18  109 N / C
2
2
12
4R  0 4 (0.07) (8.85  10 )
Half of the maximum field strength is;
E half 
iv.
0
Q
E (4R 2 ) 
ii.
Q


1
9.18  109  4.59  109 N / C
2
Because the field is uniformly distributed, the following are constants and do not change;
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k
v.
Q
40
The only variable therefore, is the distance R;
E max 7cm

E half
R
9.18  109 7cm

; R  3.5cm
R
4.59  109
QUESTION NINE
a) Use Gauss’s law to derive an expression for the capacitance of parallel plate capacitor,
whose plates have an area A and separated by a distance d. The dielectric between the plates
has a relative permittivity 𝜀 r.
b) Derive the integral form of Gauss’s law and use it to determine the electric field 24cm away
from the center of an infinite line of charge that contains a charge density of 862pC/m.
QUESTION TEN
a) Two small identical conducting spheres have charges of 2 nC and -1 nC respectively. When
they are separated by a distance of 4cm, find the magnitude of the force between them. If
they are brought into contact and the then separated by 4cm, find the force between them.
b) Consider a point located in the X-Y plane (Fig. 1) at a distance h from a straight wire of
length, Lo lying along y = 0, with a linear uniform charge density of λ C/m. Given that  1
and  2 are angles of, elevation from left end and right end of the wire respectively and that
the wire’s left end starts at x = 0, then;
a) Show that point P,  1 ,  2 and h are related to the wire of Lo meters long.
b) Show by derivation, clearly showing and explaining your procedure that;
Ey 

(cos 1  cos 2 )
40 h
Where Ey = Y-component of the field intensity at point P
ξ0 = permittivity of free space
c) Calculate the overall electric field intensity at point P, where Lo = 190m,  1 =  2 =
65°, λ = 15.6 x 10-2C/m.
Figure 1
END
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