UNIVERSITY OF BOTSWANA

Department of Physics

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PHY 241: Advanced Electricity & Magnetism

Test 1 11/03/2024

SOLUTION GUIDE

Section A
→
kq
A1. We know E = 2 (…but sign must be determined by determining field direction)
r

( )
−6
9 8.5 ×10 7 −1
Now E1 (x=6)=9.0× 10 2
=8.5 × 10 N C
(0.03)

( )
−6
9 21× 10 7 −1
and E2 (x=6)=9.0× 10 2
=21.0 ×10 N C
(0.03)
Enet ( x=6 )=( 8.5 × 107 +21× 107 )=2.95× 108 NC −1∨3.0 ×10 8 N C−1
Thus ⃗

A2.
P= |q|a = |q|l ⇒ q =
( 3 . 4×19−30 Cm
−10
1. 00×10 m
= 3 . 4×10−20 C
)
A3. Curved rod: symmetry shows that the x-component of dE cancel out because dE x+ is equal
and opposite to dEx_. Thus, net x-components of E is zero. dEy+ and dEy_ are in the same
direction, so they add up. Therefore, total e-field is in the y-direction. For the straight rod, the y-
component of e-field cancel out, and the net e-field is in the x-direction.

A4. For a solid conducting sphere, excess charge resides on the surface.
Qenc
Φ= ∮ ⃗
E⋅⃗
dA =
From Gauss’s law: ε o However, there is no charge enclosed by Gn sphere

of radius r (i.e. Qenc = 0). ⸫

⃗E = 0
A5. Total flux through the cube is ΦE = Q/ε0 and symmetry tell that the same amount of flux goes
through each of the six faces of the cube. Hence,

Section B

B1.
(a) To evaluate the E-field of the uniformly charged rod (wire) of length L at point P (distance a
from end B) along the axis of the rod.

Because of the uniform charge distribution on the rod, if the charge Q is divided by the rod
length L, we get the linear charge density  = Q/L [Cm-1].
An extremely tiny segment of length ‘dx’ therefore has a charge equal to dq = dx on it.
Elemental charge ‘dQ’ will produce elemental e-field at point P, given by
k e dQ k e λdx
dE= 2
= 2
(directed to the right)
r x
Summing up along the entire length of the rod from point P (but noting that the right end of the
rod is distance ‘a’ while and the furthest end is distance (a+
L) from point P), we can allow x to vary from a to (a+L), hence e-filed becomes:
a+ L

( )
k e λdx −1
a +L
E p =∫ dE= ∫ 2
=k e λ
a x x a

( a (a+ L) )
¿k λ −
1
e
1

( )( )
L ke Q
¿ke λ =
a (a+ L) a(a+ L)
where Q = L

(b) Given a thin rod of length l (with uniform charge per unit length λ) lying along the x-axis,
we want to show that the electric field
at P, a distance y from the rod along its
bisector, has
no x-component and is given by:
2 k e λ sin θo
E=
y
We know elemental E-field is given by:
k λ dx ^
^ = e ¿
¿ r ¿
2 2)
k e dQ (x + y
⃗
dE = r ¿
r2

Note: dE has components dEx and dEy; where

dE x =dE sin θ and dE y =dE cosθ

By symmetry x
E =∫ dE x=0 ; Thus E-field is only due to vertical components (integrated over
length l)
E=E y =∫ +l / 2 +l / 2 k λ dx ¿
e
−l / ¿+l / 2 dE = ∫ dE cos θ= ∫ cos θ
y 2+ y2)
−l / 2 −l / 2 (x

y y x
cos θ = = sin θ =
But
r √( x + y 2 )
2
and
2
r ; where r = ( x + y )
2
√
+l/2 +l/2
k e λ dx y dx
⇒E = ∫ 2 2 2 2 1/2
=k e λy ∫ 2 2 3/2
−l/2 ( x + y ) ( x + y ) −l/2 ( x + y )
dx x
∫ ( x 2+ y 2)3 /2 =
Using std integral: y 2 √ ( y 2 +x 2 ) so that:

[ ] [√ ] []
+l/ 2 +l/ 2 +θ
x k e λy x ke λ x 2 ke λ
E=k e λy = = = ( sin θ )θ0
0

y √ ( y + x ) −l /2 y r y
2 2 2 2 2 2
y y + x −l /2 −θ

(where we express the length l in terms of θ, running from 0 to θ)

2 k e λ sinθ 0 →
∴ E=
y
………………………………………………………………………………………………………

B2.
(a)
i) Problem given in class assignment (assignment #2):

Let the charge be q, so that total charge (enclosed in the box) is Qenc = (q + 2q + 4q + 8q + 16q) = 31q
From Gauss’s:
 Qenc
∅ E=∮ ⃗
E .⃗
dA=
7 2 −1
=4.8 ×10 Nm C
ϵo
Now; Q =ϵ E=( ( 8.85× 10−12 C 2 N −1 m−2 )∗(4.8 ×107 Nm2 C−1) ) =4.25 ×10−4 C
enc o

Thus 31 q=4.25 ×10−4 C

−4
4.25× 10 C −5
¿> q= =1.37 ×10 C
31
(b)
The magnitude of the e-field due to the segment containing elemental charge dQ at point P is:
kdQ dQ
dE= 2 =k 2 2 NB: E P=E x + E y
r (x +a )
dQ x kxdQ
For x-component: d E x =dEcosθ=k = 2 2 3/ 2
(x + a ) r (x +a )
2 2

kx kxQ
¿> E x =∫ d E x = 2 2 3/ 2 ∫
dQ= 2 2 3 /2 i^ and
(x +a ) (x + a )
dQ a kadQ
For y-component: d E y =dEsinθ=k 2 2
r
= 2 2 3/ 2
(x +a ) (x +a )
ka kaQ ^
¿> E y =∫ d E y = 2 3 /2 ∫
2
dQ= 2 2 3/ 2 − j
(x + a ) ( x +a )
kxQ ^ kaQ ^
Now ⃗
E p = 2 2 3 / 2 i+ 2 2 3 /2
−j
(x + a ) (x +a )
kQ
¿ 2 2 3/ 2
( x i^ +a −
^j )
(x +a )

(c) Given Q = 2×10-9 C, x = 10×10-2 m and a = 2×10-3 m then,

( 9.0 ×109 N m2 C−2 )∗( 2 ×10−9 C )
Ep= 3
( ( 10 ×10−2 )+ ( 2× 10−3 ) )
(( 10 × 10−2 )2 + ( 2 ×10 ) )
−3 2 2

= 1834.89895 or 1835 NC-1

………………………………………………………………………………………………………

B3.
(a)
k e dQx
dE =
( x 2 +r 2 )3/ 2 ;
i) Recall the E-field for the ring given by:
R R
k e x 0 2 πrσ dr 2 rdr
E disc= ∫ dEring =∫ 2 2 3/ 2
=k e x 0 σπ ∫ 2 2 3/ 2
Now disc 0 ( x 0 +r ) 0 ( x 0 +r )
 Suppose we express the variables in terms of X:
Let X = (x2+r2)  dX = 2rdr

{ }
n +1
X
Now; Ed =k e πσx ∫ X −3/ 2 dX =k e πσx ; n = -3/2 and n+1 = -1/2
n+1
Ed =k e πσx ¿ ¿
or Ed =2 k e σπ ¿
ii) Now, for xo >> R, we can ignore the value for R (i.e. R is insignificant when xo → ∞).
Thus;
→
E ∞=2 k e σπ ¿
¿0

(b)
i) The charge outside the radius r effectively contributes nothing to the field, whereas charge
inside radius r acts as if it were concentrated at the center. The volume inside radius r is
4 πr 3 /3
, so field inside the [smaller] sphere is radial and has magnitude:
( 4 πr 3 /3 ) ρ ρr
Einside= =
4 πε o r 2 3 εo
(Note: E ∝ r)
ii) For r ≥ R, the field is the same as if all charge was concentrated at the center of the
3
4π R
sphere. Since the volume of the sphere is , the field is therefore radial. Total
3
charge enclosed is;
4 πR 3 Qr
Q= ρ Q=
3 4 πε o R3
, which can be written as
From Gauss’s law, E can be written as:
Qenc
∮ E ⋅dA=E ∫ dA=¿ E ( 4 π r 2 )= εo
¿

ρ( 4 πR3 /3 ) ρR 3
⇒ E outside= =
4 πε o r 2 3 εo r 2
(Note: E ∝ r-2)
iii) The field increases linearly with r inside the sphere; the r 3 growth of the effective charge
outweighs the 1/r2 effect from the increasing distance. And the field decreases like 1/r 2 outside
the sphere. Note that E(r) is continuous at r = R, where it takes on the value ρR/3o.
NB: The answer above assumes a “conducting” sphere. Since the question did not specify
whether it is ‘non-conducting’ or conducting, students may also approach it as a “non-
conductor”.
Assuming a non-conducting spherical charge:

i) For r < R (inside), the charge outside the radius r effectively contributes nothing to the field,
whereas charge inside radius r acts as if it were concentrated at the center. The spherical
4 3
πr
3
volume inside radius r is , so that charge enclosed by smaller sphere is

( ) ( )
3
'' 4 3 Q 4 3 r
Qenc =ρ V =ρ π r ∨ πr = 3Q
3
(4
3
πR )
3 3 R

Qenc 3
r Q
So that from Gass’s law: ∮ E ⋅dA= → E ( 4 π r ) =( 3 )
2
εo R ϵo
Qr kQr
→ Ei = 3
= 3
4 π εo R R

One can also see that field inside the smaller sphere has magnitude:
3
(4 π r /3) ρ ρr
E¿ = =
4 π εo r2 3 εo

ii) For r > R (outside), the field is the same as if all charge was concentrated at the center of the
4 3
πr
3
sphere. Since the volume of the sphere is , the E-field is therefore radial. Total charge
enclosed is:
 4 πR 3 Q=
Qr
Q= ρ
3 4 πε o R3 .
, which can be written as From Gauss’s law, E can be written as:
Qenc ρ( 4 πR3 / 3) ρR 3
∮ E⋅dA = E∫ dA= E( 4 πr ) = ε
2 ⇒ E outside =
4 πε o r 2
=
3 εo r 2
o
Q kQ
One can also see (directly) that Ei i= 2
= 2
4 π ϵo r r

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