CHAPTER 23
ELECTRIC POTENTIAL
•
Potential difference and electric field
•
Potential difference between two parallel plates
•
Potential due to a single point charge
EFM08AN1.MOV
•
Potential due to a collection of charges
• Work done bringing charges together
•
Potential for continuous charge distributions
• Charged, hollow sphere
• Uniformly charged ring
When a system, e.g., you, a compressed spring, an
•
Equipotential surfaces
the same as the loss of potential energy:
•
Electrostatic potential energy
Work done and potential energy.
electric field, does (positive) work, it loses potential
energy and the amount of work done by the system is
δW = −δU.
Work done and potential energy ...
a
!
δℓ
b
!
!
F = mg
!
g
!
δℓ
! + !
F = qE
+
!
δℓ
!
E
- - - - - - - - - Let’s look at the similarity between electric and
gravitational fields. The work done by the g-field in
moving the mass from a → b is:
b!
!
δW = ∫ F • dℓ = mgδℓ ( > 0), i.e., positive work.
a
+
q#
a
b
!
F
Note: the work done by the g-field ( δW) in moving the
mass from a → b is the same as the work you do in
raising the mass from b → a.
charge from a → b is
! !
δW = F • δℓ = −δU,
energy of the charge in the field is:
! !
! !
δU = (Ub − Ua ) = − F • δ ℓ = −q #E • δ ℓ .
We define the electric potential (V) as the potential
energy per unit charge,
i.e.,
V = Uq .
#
So the potential difference between b and a
δU = Ub − Ua = −mgδℓ ( < 0), i.e., a loss.
∴δW = −δU = −(Ub − Ua ).
done by the E-field in moving the
where δU is the change in potential energy of the charge
!
!
!
in the E-field. But F = q #E , so the change in potential
The change in potential energy of the mass in the
gravitational field in moving from a → b is:
If a charge moves through a
!
!
displacement δℓ in a field E, the work
δV = Vb − Va = Vba
is given by:
! !
δV = δU q = −E • δℓ .
#
We will work problems in Cartesian (x,y,z) and polar
(r, θ,φ) coordinate systems.
[1] Cartesian coordinates:
UNITS:
!
!
E ⇒ (E x ,E y ,E z ) and δℓ ⇒ (δx, δy, δz)
! !
δV = − E • δℓ
= −(E ˆi + E ˆj + E kˆ ) • (δxˆi + δyˆj + δzkˆ )
x
i.e.,
z
δV = −(E x .δx + E y .δy + E z .δz).
∴E x = −
i.e.,
y
∂V
∂V
∂V
, Ey = − , Ez = −
∂x
∂y
∂z
!
⎛ ∂V
∂V ˆ ∂V ˆ ⎞
E = − ⎜ ˆi +
j+
k⎟ .
∂y
∂z ⎠
⎝ ∂x
These are the basic relationships between the electric
!
field, E, and the electric potential, V, in Cartesian
coordinates.
If E ⇒ N/C and
r ⇒ m
then: V ⇒ volts (V).
But, by definition:
Ex = −
dV
, etc.,
dx
then: E ⇒ V/m,
which means that N/C ≡ V/m.
(It is more usual to use V/m as the unit of electric field.)
Conventional definition of work done in an electric
field ...
DISCUSSION PROBLEM:
"
δℓ
+
q!
~10,000V
a
b
"
F
120V
The work done by the field in
moving the charge from a → b is
δW = −δU = −(Ub − Ua )
= −(q ! Vb − q ! Va ) = q ! (Va − Vb ).
(Remember, by definition ⇒ V = U q )
!
Conventionally, when a charge moves from a → b we
When you charge a balloon by friction, its electric
write the work done by the field as:
potential is ~10,000V, but it is safe to handle! And yet,
δW = q ! (Va − Vb ) = q ! Vab ,
where Vab is the potential difference between the start
a typical socket operates at a potential of 120V but will
give you a (potentially!) fatal shock.
*
What’s the difference?
*
Why is the socket more “shocking”?
point (a) and the end point (b). (Note also if the charge
was released and free to move in the field, δK = δW.)
Therefore, the work done by you in moving a charge
from a → b is:
δW = −q ! Vab = −q ! (Va − Vb ).
+σ
Example using Cartesian coordinates ...
• Potential between two parallel charges plates
!
E = E x ˆi
1
2
2
1
− ΔV +
b
d
!
ℓ
a
d
ˆj
V1
!
between a and b (displacement ℓ )
!
in a field E = E ˆi produced
x
between two parallel, infinitely
large charged plates, spaced a distance d apart. Along
the displacement, the change in potential is:
! !
dV = − E • d ℓ = −E xˆi • (dxˆi + dyˆj) = −E xdx.
b
b
a
a
∴Vb − Va = ∫ dV = −E x ∫ dx = −E x x = −
σ
∴Vb = Va − x
ε#
σ
x
ε#
"
E = E x ˆi
2
V2
d
V
d
ΔV
σ
x
ε!
For the pair of plates
Vb = Va −
(V2 − V1 ) = ΔV = −
V1
Find the potential difference
!
ℓ = xˆi + yˆj
ˆi
−σ
1
i.e., V1 > V2 .
σ
d,
ε!
x
V2
• ΔV is independent of y, it depends only on σ and d.
Thus, ΔV is the same between any point on plate 1 and
any point on plate 2 . This means that the potential is
constant over an infinitely charged plate.
•
The work done by the field in moving a charge q
from a → b is:
δW = q(V1 − V2 ) > 0,
so a +ve charge moves from a position of higher
potential ( V1) to lower potential ( V2).
Question 23.1: Shown here is a plot of electric potential
versus distance where there is an electric field. Where is
V
a
the magnitude of the electric field greatest?
x
e
b
c
V
a
b
c
x
e
d
By definition: E x = −
d
dV
, so E x is a maximum at d and e.
dx
+σ
(a,b) Since the field would move a
−σ
+ve charge from the +σ plate to the
1
e−
Question 23.2: The facing surfaces of two large parallel
conducting plates separetd by 10.0 cm have uniform
surface charge densities +σ and −σ. The difference in
potential between the plates is 500V. (a) Which plate has
2
−σ plate, the +σ plate is at the higher
potential.
10 cm
ΔV = (V1 − V2 ) = 500V = −E.d,
∴ E = ΔV d = 500 0.1 = 5000 V/m.
(c) Work done by the field is W2→1 = qV21.
the higher potential? (b) What is the size of the electric
But V21 = (V2 − V1 ) = −500 V.
field between the plates? (c) An electron is released from
∴W2→1 = −1.6 × 10−19 × (−500) = 8.0 × 10−17 J.
rest next to the negatively charged plate. What is the
work done by the electric field as the electron moves
from the negatively charged plate to the positively
charged plate? (d) What is the change in potential energy
of the electron as it moves from one plate to the other?
(e) What is its linetic energy when it reaches the
positively charged plate?
(d) The change in potential energy of the electron:
ΔU = U1 − U2 = qV1 − qV2 = q(V1 − V2 )
= −1.6 ×10 −19 × 500 = −8.0 ×10 −17 J.
(e) Mechanical energy is conserved:
∴ΔK = −ΔU = 8.0 × 10−17 J.
0
1
2ΔK
ΔK = m(v 2 − v !2 ). ∴v =
= 1.33 × 107 m/s.
2
m
Note also, from the work - kinetic energy theorem:
ΔK = W2→1 = qV21.
From earlier, the components of the electric field are
related to the electric potential in the (x, y, z) coordinate
system by the relationships:
∂V
∂V
Ex = −
Ey = −
∂x
∂y
Ez = −
Since V = (4xz − 5y + 3z 2 ), we have
Question 23.3: The electric field in a certain region of 3dimensional space varies as:
V = (4xz − 5y + 3z 2 ) volts.
What is the electric field at (2, −1,3), where all distance
are in meters.
Ex = −
∂V
= −4z,
∂x
Ey = −
∂V
= 5,
∂y
Ez = −
∂V
= −4x − 6z.
∂z
!
So, at r = (x, y,z) ⇒ (2, −1,3) m,
E x = −4z = −12 V/m,
E y = 5 V/m,
E z = −4x − 6z = −26 V/m.
!
∴ E = −12ˆi + 5ˆj − 26 kˆ V/m.
(
)
∂V
.
∂z
Example ... electric potential for a point charge:
[2] Polar coordinates
If the electric field has radial symmetry, i.e., it depends
!
only on r , e.g., a point charge, then
!
Q
E
(r) ⇒ E r ˆr = k 2 ˆr .
ˆr
r
!
!
d
r
+
For a radial displacement dr (in
!
But dr // ˆr ,
the ˆr direction):
!
!
!
dV(r) = − E(r) • d r = −E r ˆr • d r .
=1
!
!
∴ ˆr • d r = ˆr d r cos0 = dr.
Therefore, potential difference between radii r2 and r1 is
r2
V21 = V(r2 ) − V(r1 ) = − ∫ E r dr.
r1
But dV(r) = −E r dr, so the radial electric field is
dV(r)
Er = −
.
dr
Again, we have simple relations between the electric field
E r and the electric potential V(r).
The electric field of a point
Q
+ a• d!r
•
b
ˆr
charge is:
!
Q
E(r) = k 2 ˆr .
r
!
For a small displacement dr in the radial direction ( ˆr ),
the change in potential is:
!
Q
Q
!
!
dV(r) = − E(r) • d r = −k 2 ˆr • d r = −k 2 dr.
r
r
Q
Q
∴V(r) = −k ∫ 2 dr = k + V" ,
r
r
where V" is an integration constant. If we define the
electric potential at infinity as zero, i.e., V(r → ∞) = 0,
then V" = 0. So, the absolute electric potential at r is
Q
V(r) = k ,
r
⎡1 1⎤
∴Vb − Va = kQ ⎢ − ⎥
(i.e., < 0 if Q is +ve)
⎣ rb ra ⎦
Go from a → b by different routes. The potential at any
point a distance r from a
“equipotentials”
a
Electric
potential (V)
point charge is:
Q
V(r) = k .
r
x
b
Since ra = rx , Va = Vx , so
the potential difference
Vab = Vxb ,
y
i.e., the potential difference between two points does not
depend on the path between them only the potentials at
the end points. The work done by you in moving a
x
V= k
Q
r
charge q from a → b by the two different routes is:
The electric potential for a positive charge. If the charge
[1] Wa →b = −q(Va − Vb ) = −qVab .
[2] Wa →x→b = [ −q(Va − Vx )] + [ −q(Vx − Vb )]
is negative, the potential looks like a “hole” rather than a
= −q(Vx − Vb ) = −qVxb .
“hill”.
Note that as x (and y) → ±∞, V → 0.
But Vab = Vxb .
∴Wa → b = Wa →x→ b .
So, the work done by you ( = −qΔV) in moving a charge
from one point to another does not depend on the path
... only on ΔV.
The electric force is a conservative force. Here are two
reasons ...
[1] The work done by the electric force in moving a
charge from one point to another is independent of the
path ... a property of a conservative force.
[2] We can write a potential (energy) function, which
Question 23.4: Is the electric force a conservative or
non-conservative force?
can only be done for conservative forces.
Q
For a point charge V(r) = k . Therefore, the potential
r
energy of a test charge q ! in the field due to a point
charge Q is
qQ
U(r) = q ! V(r) = k ! .
r
i.e., a simple function of r.
No matter what the source, the electric force is a
conservative force.
+
Question 23.5:
r2
+
2
+
+
2
r1
+
+
1
A proton is fired towards a helium nucleus. If the speed
of the proton at point 1 is v1 and its speed at point 2 is
v 2, which of the following statements is correct?
A: v 2 < v1.
B: v 2 > v1.
C: v 2 = v1.
1
Use the conservation of mechanical energy:
K2 + U2 = K1 + U1.
The potential energy of the proton (with charge q) in the
field of the helium nucleus at 1 is U1 = qV1, and its
potential energy at 2 is U2 = qV2 .
Since V = k
Q
and r2 < r1, then V2 > V1.
r
∴U2 > U1.
So,
K2 < K1,
i.e.,
v 2 < v1,
which means A is the correct choice.
A
+2µC
3m
3m
3m
B
+2µC
Question 23.6: Points A, B and C are at the vertices of
an equilateral triangle whose sides are 3.00 m long.
Point charges of +2.00 µC are fixed at A and B.
(a) What is the electric potential at point C?
(b) How much work is required to move a +5.00 µC
V= k
C
Q
r
(a) Potential at C is due to both QA and QB ( = Q)
Q
Q
Q
VC = k A + k B = 2k ,
r
r
r
where r is the length of the sides.
∴VC
2 × 10 −6 )
(
= 2 × (9 × 10 ) ×
= 1.2 × 10 4 volts.
9
3
point charge from infinity to the point C?
(c) How much additional work is required to move the
(b) Work done by you in bringing a charge q ! from ∞ to
+5.00 µC charge from C to the midpoint of side AB?
the point C is:
W = −q ! ΔV = −q ! (V∞ − VC ) = q ! VC
(
) (
= 5 ×10 −6 × 1.2 ×10 4
= 6.0 × 10−2 J.
)
A
Potential due to a spherical shell of charge
• on a hollow or solid conducting sphere ...
+2µC
+ +R +
+
+
+
+
+
D×
B
+2µC
C
(c) The extra work done in moving the charge from C to
D is
δWC→D = −q ! ( VC − VD ) = q ! (VD − VC )
Q
Q
Q
But
VD = k A + k B = 4k
r
r
r
2
2
Q
From earlier, VC = 2k .
r
⎛ Q
Q⎞
Q
∴δWC→D = q ! ⎜ 4k − 2k ⎟ = 2q ! k
⎝ r
r⎠
r
( )
(
= 2 × 5 × 10
−6
( )
2 × 10 −6 )
(
) × (9 × 10 ) ×
9
3
Q σ
Es = k 2 =
ε!
R
Q
4πR 2
E(r)
E=0
From earlier:
σ⇒
E=−
dV
so
dr
Q
E=k 2
r
r
dV = −E.dr,
Q
Q
∴V(r > R) = − ∫ E.dr = −k ∫ 2 dr = k
r
r
r >R
V(r)
Vs = k
Q
R
V= k
= 6.0 × 10−2 J.
Q
r
r
But what about inside the sphere?
Potential due to a spherical shell of charge
• on a hollow or solid conducting sphere ...
Q σ
Es = k 2 =
ε!
R
TWO POINTS:
[1]
E(r)
E=0
[2]
Q
E=k 2
r
r
V is constant inside a conducting sphere
(i.e., the same as at the surface).
Q
Q
At the surface: Vs = k and Es = k 2 .
R
R
V
∴Vs = Es .R or Es = s .
R
As the charge Q on the sphere increases, so do
⎛
⎞
Vs = k Q R and Es ⎜ = k Q 2 ⎟ .
⎝
R ⎠
Inside the sphere, i.e., for r < R, dV = −E.dr = 0
(
∴V(r < R) = constant.
If V is constant inside sphere, no work is done in
moving a charge anywhere inside the sphere. Then
W = −q !ΔV = 0, i.e., ΔV = 0.
Q
∴V(r < R) = k ⇒ constant.
R
)
Under “normal conditions” the maximum electric field
obtainable in air before breakdown is
Emax ~ 3 × 106 V/m.
This sets a maximum potential and a maximum charge for
a spherical conductor (radius R):
V(r)
Vs = k
Q
R
i.e.,
Vmax = E max .R ~ 3 × 106 R volts.
Q
3 × 106 2
Since, E = k 2 , then Qmax ~
R Coulombs.
k
R
Q
V= k
r
r
Larger R means larger Vmax and Qmax before breakdown.
Assume the Earth is a sphere. The potential at the surface
of a sphere is:
Vs = Es .R.
For the Earth: Vs = Es .R = (200 V/m ) × (6400 × 10 3 m)
Question 23.7: We know that the electric field near the
= 1.28 × 109 volts!
Earth’s surface is ~200 V/m. If the Earth has a radius of
about 6400 km, what is the Earth’s electric potential?
Note: in question 22.4, we found that the net charge on
the Earth is negative ... −9.11 × 105 C ... so the electric
potential at the surface is
Vs = k
Q
R
<0
Also, we can now show why charges “pile-up” at sharp
points on a charged conductor ...
Region 1 ⇒ Q1
radius R1
σ1
DISCUSSION PROBLEM:
If the electric potential of the Earth is so large, how come
we aren’t fried to a crisp when standing barefoot on the
Earth’s surface?
Region 2 ⇒ Q2
radius R 2
σ2
The potential inside the conductor is constant
Q
Q
∴k 1 = k 2 .
R1
R2
But
Q1 ≈ 4πR12σ1 and Q2 ≈ 4πR 22σ2
so
i.e.,
R1σ1 ≈ R 2σ 2
R
σ2 ≈ 1 σ1.
∴σ2 > σ1.
R2
σ
σ
Also, E1s = 1 and E2s = 2 . ∴E2s > E1s .
ε!
ε!
Therefore, the charge density and the surface electric
field are greater at “points”.
R2
R3
R3
R1
1
2
3
We know that V = EsR. Therefore, the maximum
1
2
3
Question 23.8: These objects are charged to their
maximum potential (voltage) before breakdown of the
potential (when breakdown occurs) Vmax ∝ R. Since
R 3 > R1 > R 2 , the order from largest maximum potential
to smallest is:
3 :1: 2
surrounding air. Rank them in order from the one with
the largest voltage to the smallest voltage.
Note that at the sharp end
of #2 the electric field is
greatest; that’s where
breakdown first occurs.
Take two charged spheres A ( VA) and B ( VB) with
VA > VB ... connect them together by a conducting wire.
Q
VA = k A
RA
q
A
+
B
The force on charge q is: F = qE = −q
so
Q
VB = k B
RB
dV
, but VA > VB
dℓ
dV
< 0, therefore, +ve charges move from A to B, i.e.,
dℓ
from high potential ( VA) to low potential ( VB).
Question 23.9: A spherical conductor of radius
R 1 = 24.0 cm is charged to 20.0 kV. When it is
connected to by a long, very thin conducting wire to a
second conducting sphere a great distance away, its
V
∴QA and VA decrease
second sphere?
+
∴QB and VB increase
As charges move from A to B, VA decreases and VB
increases. When VA = VB, charges stop moving because
when ΔV = 0 then F = 0.
potential drops to 12.0 kV. What is the radius, R 2, of the
V1 = 20kV → V1′ = 12kV
V2 = 0 → V2′ = 12kV
R1
R2
1
2
After they are connected we have:
q
q
V1′ = k 1 = V2′ = k 2 = 12 × 10 3 V,
R1
R2
R
R
i.e., q 1 = (12 × 10 ) 1 and q 2 = (12 × 10 3 ) 2 .
k
k
3
If sphere #2 is initially uncharged ( V2 = 0). Since charge
is conserved, before they are connected we have
(q + q 2 )
V1 = k 1
= 20 × 103 V,
R1
(
i.e., q 1 + q 2 = 20 × 10 3
∴(12 × 10 3 )
) Rk1 .
R1
R
R
+ (12 × 10 3 ) 2 = (20 × 10 3 ) 1 ,
k
k
k
i.e., (12 × 10 3 )R 2 = (8 × 10 3 )R 1.
8
2
∴R 2 = R 1 = × 24.0 cm = 16.0 cm.
12
3
q2
q1
R2
R1
R 2 = 2R1
Question 23.10: Initially, two, well separated conducting
spheres are given the same charge (q). They are then
connected to each other with a conducting wire. If
R 2 = 2R1, what are the final charges, q1 and q 2, and
charge densities, σ1 and σ 2 , on each sphere?
Electrostatic potential due to a uniformly charged ring:
q2
q1
dQ
+ +
R2
R1
a
R 2 = 2R1
+
+
If they have the same initial charge, the potential of #1
( V1) will be greater than the potential of #2 ( V2), because
R 1 < R 2 . After connecting the two spheres together,
r = x2 + a 2
+
P
×
dV
x
+
charges will flow from from #1 to #2 until the potentials
The potential at the point P due to the element of charge
are the same, i.e., V1 = V2 .
q
q
∴k 1 = k 2 ,
R1
R2
dQ is:
(
i.e.,
q1 q 2
=
.
R1 2R1
∴q 2 = 2q1.
)
The surface area 4πR 2 of #2 is four times the surface
area of #1, i.e., A 2 = 4A1.
q
2q
1
∴σ 2 = 2 = 1 = σ1.
A 2 4A1 2
dV = k
dQ
.
r
∴V(x) = ∫ k
dQ
.
r
But r = x 2 + a 2 ⇒ constant.
∴V(x) =
k
x2 + a
dQ = k
2∫
where Q is the total charge on the ring.
Q
x2 + a 2
Electrostatic potential due to a uniformly charged ring:
dQ
+ +
r = x2 + a 2
a
+
+
+
Q
a
x
a
x
P
×
+
V(x) = k
V= k
P
×
dV
Question 23.11: What difference would it make, if any,
Q
if the charge Q were NOT uniformly nor symmetrically
x2 + a 2
distributed around the ring?
V= k
Q
x
A: The potential at any point P on the axis would be
larger.
x
-2
0
2
4
Q
When x = 0, V(x) = k
a
Q
When x >> a, V(x) ⇒ k ,
x
i.e., the ring looks like a point charge.
6
B: The potential at any point P on the axis would be
a
smaller.
C: The potential depends on the actual distribution.
D: It would make no difference.
EQUIPOTENTIAL SURFACES ...
dQ ⇒ Q
r
a
An equipotential surface is a 3-dimensional surface over
P
×
V
which the potential is constant.
• for an isolated positive point charge.
V = +30V
Take an extreme scenario, i.e., with all the charge
concentrated at one point on the ring. The total potential
V = +50V
at P due to dQ is
V= k
dQ
Q
⇒k ,
r
r
i.e., the same as before!
Since potential is a scalar quantity it makes no difference
if the charge is concentrated in one region or distributed
+Q
V = +70V
!
Q
Q
From earlier: E(r) = k 2 ˆr and V = k .
r
r
around the ring; it only depends on r, the distance of P
So, the equipotential surfaces for an isolated point charge
from the ring.
are spheres centered on the charge. Equipotential
!
surfaces are always perpendicular to the E lines. Also,
Therefore, the answer is D.
no work is done moving a charge around an equipotential
surface (since ΔV = 0).
EQUIPOTENTIAL SURFACES ...
• for two equal and opposite point charges
V = −50V
V = −30V
“equipotentials”
V = +30V
V
V = +50V
x 0
+Q
−Q
0
V = +70V
V = −70V
y
0
V = 0V
The equipotential surfaces are closed surfaces
!
perpendicular to the E lines at all points. The infinite
Three dimensional plot of the electric potential for two
plane equidistant from the charges and perpendicular to
equal and opposite charges.
the line joining them is an equipotential surface of zero
potential. Note that very close to each charge, the
equipotential surfaces are spherical.
EQUIPOTENTIAL SURFACES ...
EQUIPOTENTIAL SURFACES ...
• for two oppositely charged parallel plates
• for two equal point charges
V
V = +30V
400
+++++++++++
E
V = +50V
+Q
+Q
V = +70V
0
- - - - - - - - - - -
!
The equipotential surfaces are perpendicular to the E
Electric field lines (green) and equipotential surfaces
field lines. Very close to the charges the equipotential
(red) for two parallel plates with a potential difference of
surfaces are approximately spherical but as the distance
400V between them. Note, everywhere the
increases the surfaces become distorted, but are
equipotential surfaces are perpendicular to the electric
symmetrical.
field lines and between the plates the equipotential
surfaces are planar (i.e., flat).
Question 23.12: Several equipotentials are shown labelled
in volts. The spacing of the grid is 1.00cm. What is the
magnitude and direction of the electric field at
(a) X,
(b) Y?
y
X
Y
Equipotential
Using the property that equipotential surfaces are
!
perpendicular to the E field lines, enables us to sketch
15V
the equipotential surfaces for non-spherical charged
objects like that shown above.
x
20V
10V
5V
1 cm
0V
1 cm
y
X
Y
y
X
x
20V
Y
15V
10V
5V
1 cm
0V
1 cm
The electric fields at X and Y are parallel to the x-axis.
!
∂V ˆ
∴E = −
i.
∂x
!
ΔV
(15 − 5) ˆ
10 ˆ
(a) at X: E = − ! = −
i=−
i = −500ˆi V/m.
Δx
0.02
0.02
The negative sign means the electric field is in the −x
!
direction. (Note also that E is always directed from higher
potential to lower potential ... it’s the direction a +ve
charge would move.)
x
20V
15V
10V
5V
1 cm
0V
1 cm
(b) at Y:
!
ΔV
(5 − 15) 10
E=− ! =−
=
= +500 V/m.
Δx
0.02
0.02
The positive sign means the electric field is in the +x
direction.
Electrostatic potential energy of a collection of charges:
q1
A
r12
B
Now bring in a charge q 3 to the point C.
q1
A
From
∞
q2
r12
r13
B
q2
r23
C
From
∞
q3
The potential energy of an ensemble of charges is
The work done in bringing charge q 3 from ∞ to C is:
simply equal to the work done in bringing the charges
together. Assume charge q1 is in position A and bring in
W3 = −q 3(V∞ − VC ),
where VC is the potential at C due to the charges at A
charge q 2 to B from ∞. The potential at B due to a
and B.
charge q1 at A is:
q
VB = k 1 .
r12
So, the work done (by you) in bringing a charge q 2
from ∞ to B is:
W2 = −q 2ΔV = −q 2 (V∞ − VB ) = −q 2 (0 − VB )
qq
= k 1 2.
r12
⎛
q
q ⎞
qq
q q
∴W3 = −q 3⎜ 0 − (k 1 + k 2 )⎟ = k 1 3 + k 2 3 .
⎝
r13
r23 ⎠
r13
r23
So, the total work done = W2 + W3
qq
qq
q q
= k 1 2 + k 1 3 + k 2 3.
r12
r13
r23
This is the electrostatic potential energy, U, of the three
charges at A, B and C, i.e., the total work done
assembling the charges.
r12
q1
r13
r23
q2
q3
So, the electrostatic potential energy, U, of a system of
point charges is the total work done to assemble the
charges from ∞ to their final positions. Now,
qq
qq
q q
U= k 1 2 +k 1 3 +k 2 3
r12
r13
r23
1 ⎛ q
q ⎞ 1 ⎛ q
q ⎞
= q1 ⎜ k 2 + k 3 ⎟ + q 2 ⎜ k 3 + k 1 ⎟
2 ⎝ r12
r13 ⎠ 2 ⎝ r23
r12 ⎠
1 ⎛ q
q ⎞
+ q3⎜ k 1 + k 2 ⎟
2 ⎝ r13
r23 ⎠
=
1 N
∑ q i Vi ,
2 i=1
where Vi is the potential at the position of the ith charge
due to all the other charges.
Question 23.12: Point charges, q 1 = q 2 = −4.20 µC and
q 3 = +4.20 µC are fixed at the vertices of an equilateral
triangle, whose sides are 2.50 m long. What is the
electrostatic potential of this ensemble of charges?
Electrostatic potential of a charged sphere
+4.2µC
2.5m
−4.2µC
2.5m
The electrostatic potential energy of a charged conducting
2.5m −4.2µC
sphere is equal to the amount of work we do in putting the
The electrostatic potential energy is: U =
1 3
∑q V
2 i=1 i i
1 ⎡ q
q ⎤ 1 ⎡ q
q ⎤ 1 ⎡ q
q ⎤
= q 1 ⎢k 2 + k 3 ⎥ + q 2 ⎢ k 3 + k 1 ⎥ + q 3 ⎢k 1 + k 2 ⎥
2 ⎣ r12
r13 ⎦ 2 ⎣ r23
r12 ⎦ 2 ⎣ r13
r23 ⎦
⎡(−4.2 × 10 −6 ) (+4.2 × 10 −6 ) ⎤
1
= (−4.2 × 10 −6 ) × (9 × 10 9 )⎢
+
⎥
2
2.5
2.5
⎣
⎦
⎡(+4.2 × 10 −6 ) (−4.2 × 10 −6 )⎤
1
+ (−4.2 × 10 −6 ) × (9 × 10 9 )⎢
+
⎥
2
2.5
2.5
⎣
⎦
⎡(−4.2 × 10 −6 ) (−4.2 × 10 −6 )⎤
1
+ (+4.2 × 10 −6 ) × (9 × 10 9 )⎢
+
⎥
2
2.5
2.5
⎣
⎦
= (+4.2 × 10 −6 ) × (9 × 10 9 )
= −0.0635J
?
What does the negative sign mean?
(−4.2 × 10 −6 )
2.5
dq
q
R
+
charge onto the sphere. If
the sphere already has a
charge q, the work done in
bringing a charge dq from ∞
onto the sphere is dW = −dq(V∞ − V), where V is the
q
potential of the sphere. But V∞ = 0 and V = k R ,
q
∴dW = k R dq.
( )
The total work done to charge the sphere from 0 → Q is:
Q
( )
Q
k ⎡q 2 ⎤
1 Q2
1
W = ∫ k q R dq =
=
k R = QV,
2
⎥⎦ 0 2
R ⎢⎣
2
0
where V ( = k Q R ) is the potential of the fully charged
sphere.
1
∴U = W = QV.
2
Q
Question 23.13: An isolated spherical conductor with a
(a) We’ve just shown that the
electrostatic potential energy of a
1
charged sphere is: U = QV.
2
Q
VR
But, V = k , i.e., Q =
.
R
k
R
Substituting for Q, we find
1 VR ⎞
V2R
U= ⎛
V=
2⎝ k ⎠
2k
radius of 10.0 cm is charged to 2.00 kV.
(a) What is the electrostatic potential energy of the sphere?
(b) What is the charge on the surface of the sphere?
=
(2 × 103 )2 × 0.1
= 2.22 × 10 −5 J.
9
2 × 9 ×10
This is the amount of work we do in charging the sphere
to a potential of 2kV.
(b) The charge on the sphere is (from above):
RV 0.1× 2 × 103
Q=
=
= 2.22 × 10−8 C ( 22.2 nC).
9
k
9 × 10