ENERGY IN ELECTROSTATIC
FIELDS
LO 2.7
WORK, POTENTIAL & ENERGY
LO 2.7 Identify Energy in Electrostatic Fields.
•
Explain the concept of energy exchanges
in the electric field of a system of charges.
•
Explain the concept energy expended to
build up a system of charges.
•
Perform calculation related to energy in
electrostatic fields.
WORK DONE IN MOVING A
POINT CHARGE
WORK DONE IN MOVING A POINT
CHARGE.
•
A charge Q experience a force F in an
electric field
E.
•
In order to maintain the charge in equilibrium
a force
F
a must be applied in opposition.
F
a=Q
E
F
= Q
E
E
F
=
F
a
F
=
Q
E
F
a
= Q
E
•
Work
= a force acting over a distance.
•
Therefore, a differential amount of work, dW
is done when the applied force
F
a
produces a differential displacement d
ℓ
of the
charge i.e move the charge
through the
distance.
WORK DONE
d
ℓ
= d
ℓ

dW =
F
a
.
d
ℓ
= Q
E .
d
ℓ
If Q is +ve & d
ℓ
is in the direction of
E
,
dW = QE dℓ <o, indicating that work was
done by the electric field.
Component forms of the differential
displacement vectors are as follows:
d
ℓ
= dx
a
x
+ dy
a
y
+ dz
a
z
(cartesian)
d
ℓ
= dr
a
r
+ rdø
a
ø
+ dz
a
z
(cylindrical)
d
ℓ
= dr
a
r
+ rdθ
a
θ
+ rsinθdø
a
ø
(spherical)
Differential displacement vectors, d
ℓ
Recall
Cartesian coordinate system
Component form of the
differential displacement
vector,
d
ℓ
= dx
a
x
+ dy
a
y
+ dz
a
z
Component form of
the differential
volume vector,
d
v
= dxdydz
Recall
Cylindrical coordinate system
d
ℓ
= dr
a
r
+ rdø
a
ø
+ dz
a
z
differential displacement
vector d
ℓ,
d
v
= rdrdødz
differential volume
vector d
v
,
Recall
Spherical coordinate system
d
ℓ
= dr
a
r
+ rdθ
a
θ
+ rsinθdø
a
ø
differential displacement
vector d
ℓ,
d
v
= r
2
sinθdrdθdø
differential volume
vector
d
v
,
Recall
An electrostatic field is given by
E
=
(x/2 + 2y)
a
x
+
2x
a
y
(V/m). Find the work
done in moving a point charge
Q = 20 μC
a)From the origin to (4,0,0) m, and
b)From (4,0,0) m to
(4,2,0) m.
WORK DONE
EXAMPLE 1
a) The first path is along the x axis i.e
from (
0
,0,0)m to (
4
,0,0) m.
So that, differential displacement,
d
ℓ
= dx
a
x
dW =
Q
E.
d
ℓ
W =
Q ∫
E.
d
ℓ
= (20 μ)∫{(x/2 + 2y)
a
x
}
.
{dx
a
x
}
= (20 μ)∫(x/2 + 2y)dx = 80 μJ
WORK DONE
SOLUTION
b) The 2
nd
path is in the
a
y
direction i.e
from (4,
0
,0)m to (4,
2
,0) m.
So that, differential displacement,
d
ℓ
= dy
a
y
dW =
Q
E.
d
ℓ
W =
Q ∫
E.
d
ℓ
= (20 μ)∫{(2x)
a
y
}
.
{dy
a
y
}
= (20 μ)∫2xdy = 320 μJ
WORK DONE
SOLUTION
WORK DONE 
Conservation
property of the electrostatic field
The
work done
in moving a point dharge
from one location, B to another, A, in a
static electric field is
independent
of the
path
taken. Thus, in term of figure below,
1
2
A
B
W
1
=
W
2
∫
1
E.
d
ℓ =
∫
2
E.
d
ℓ
Or
∫
12
E.
d
ℓ
= 0
Where the last integral is over the closed contour formed
by 1 described positively and 2 described negatively.
WORK DONE 
Conservation
property of the electrostatic field
•
Conversely, if a vector field
F
has the
property that
∫
F.
d
ℓ =
0 over every closed
contour, then the value of any line integral
of
F
is determined solely by the endpoints
of the path.
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 Fall '19
 Sabariah