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Week 3 Weekly Outline (SCS 139) • Electric Potential
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Week 3
Weekly Outline (SCS 139) • Electric Potential
Week 3
Electric Potential : Work Done by a Field
• i: The internal energy of the system is Ui
• f: The internal energy of the system is Uf
• The work done by the field in moving the charge from position i to position f is equal to qEd (that’s force times distance).
• Therefore, the change in energy WU −=∆
Week 3
Electric Potential : Work Done by a Field
• Electric potential is the energy required per unit charge
• The electric potential difference ∆V between any two points i and f in an electric filed is
qUV =
Week 3
Electric Potential : Work Done by a Field
• Because potential energy is relative. We can set the energy at infinity to be zero. The electric potential at any point is then the amount of work required to bring a 1 C charge to that point from infinity.
Week 3
Calculating the Potential from the Field
∫
∫∫
=
==
=
=
f
i
f
i
f
i
sdEq
sdEqdWW
sdEq
sdFdW
.
.
.
.
0
0
0
∫−=−
=∆
=−=∆f
iif sdE
qW
qUVVV
.00
Week 3
Electric Potential : Due to a Point Charge
If we let the potential energy at infinity to be zero
∫−=−f
iif sdEVV
.
rq
drr
q
sdEV
r
f
if
0
20
41
4
.
πε
πε
=
−=
−=
∫
∫
∞
Positive charge produces a positive electric potential and negative charge produce a negative electric potential
Week 3
Potential: Due to a Collection of Charges
• Potential from one charge is
Because potential is scalar, for a collection of charges, the potential is
Week 3
Potential: Due to an Electric Dipole
Week 3
Potential: Due to an Electric Dipole
20
20
0
00
cos4
1
cos4
4
44
rp
rdq
rrrrq
rq
rq
VVV
θπε
θπε
πε
πεπε
=
=
−=
−=
+=
+−
+−
−+
−+
Week 3
Example
• Particles of charges q1 = +5e and q2 = -15e are fixed in place with a separation of d = 24.0 cm. With V = 0 at infinity, what are the finite (a) positive and (b) negative values of x at which the net electric potential on x axis is zero?
Week 3
Solution • For x > d, there is no solution. • For 0 < x < d,
solve: x = d/4. With d = 24.0 cm, x = 6.00 cm. • For x < 0,
solve: x = d/2. With d = 24.0 cm, x = –12.0 cm
Week 3
Example • A rectangular array of charged particles
fixed in place, with a distance a = 39.0 cm and the charges shown as integer multiples of q1 = 3.4 pC and q2 = 6.0 pC. With V = 0 at infinity, what us the net electric potential at the rectangular center?
Week 3
Solution V → 0 as r → ∞, all corner particles are equidistant
from the center, and since their total charge is 2q1– 3q1+ 2q1– q1 = 0, then their contribution is zero. The net potential is due, then, to the two +4q2
particles, each of which is a distance of a/2 from the center. It is
Week 3
Potential: Due to a Line of Charge
Week 3
Potential: Due to a Line of Charge
( )( )( )[ ]( )
++=
++=
+=
=
∫
∫
ddLL
dxx
dxdx
rdqV
L
L
2/122
0
022
0
022
0
0
ln4
ln4
41
41
πελ
πελ
λπε
πε
Week 3
Example • A nonconducting rod of length L = 6.00 cm
and uniform linear charge density λ = +3.68 pC/m. Take V = 0 at infinity, what is V at point P at distance d = 8.00 cm along the rod’s perpendicular bisector?
Week 3
Solution
Week 3
Potential: Due to a Charged Disk
Week 3
Potential: Due to a Charged Disk
( )( )( )
( )
( )zRz
zR
zR
dRR
zR
dAr
dqV
R
o
R
−+=
+=
+=
+=
=
∫
∫
∫
22
0
22
0
022
0
220
0
2
'2
'4
''2
'4
4
εσεσ
πε
πσ
πε
σπε
Week 3
Example • A plastic disk of radius R with a uniform
surface charge density σ = 7.73 fC/m2 and then three quadrants of the disk are removed. Take V = 0 at infinity, what is V at point P at distance D = 25.9 cm from the origin center?
Week 3
Solution
• A disk is divided into 4 pieces. So, each quadrant provides only a quarter of the total.
Week 3
Calculating the Field from the Potential
• The component of E in any direction is the negative of the rate at which the electric potential changes with distance in that direction
zVE
yVE
xVE zyx ∂
∂−=
∂∂
−=∂∂
−= ; ;
Week 3
Example
• The electric potential at any point on the central axis of a uniformly charged disk is given by
Find the electric field.
( )zRzV −+= 22
02εσ
Week 3
Example • The electric potential at any point on the
central axis of a uniformly charged disk is given by
( )zRzV −+= 22
02εσ
