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Lecture 6-1Today
• Electric “Shielding”
• Electric Potential
Lecture 6-2 Electrostatic “Shielding” I
If you move charge q in the cavity,
the exterior electric fields and the
extreior charge distribution are not
affected.
Conducting shell electrostatically
“shields” its exterior from
changes on the inside.
• As long as the E contribution due
to all the interior charges (both in
void area and on interior surface),
measured at the outer surface
remains zero, then moving the +q
charge does not affect the exterior
surface charge distribution.
Truth:
Superposition,
NOT shielding!
q
Lecture 6-3Electrostatic Shielding II
Conducting shell electrostatically
“shields” its interior from changes
in the exterior, and vice versa.
If you now add charge Q’ to the conductor
and/or Q’’ on the outside of the conductor,
the interior electric fields do not change.
Add Q’Q’’ Q’Q’’
Truth:
Superposition,
NOT shielding!
Like active noise
cancellation
+++
+
+
+
Lecture 6-4
Electric Potential Energy of a Charge in Electric Field
• Coulomb force is conservative
=> Work done by the Coulomb
force is path independent.
• Can associate potential energy
to charge q0 at any point r in
space. ( )U r
It’s energy! A scalar measured in J (Joules)
d l
0dW q E dl= Work done by E field
0dU dW q E dl= − = − Potential energy change
of the charge q0
Lecture 6-5Electric Potential Energy of a Charge (continued)
0dW q E dl=
0
( ) ( )
r
i
U U r U i
q E dl
= −
= −
0dU dW q E dl= − = −
i is “the” reference point.
Choice of reference point
(or point of zero potential
energy) is arbitrary.
0
d l
i is often chosen to be
infinitely far ( )
Lecture 6-6
Gravitational vs Electrostatic Potential Energy
( ) ( )
b
a
U U b U a
dF l
= −
= −
a
b
qEmgGravity Coulomb
Work done by<gravity/Coulomb> force is the
decrease in potential energy.
mg l− qE l−
(if g, E uniform)
Lecture 6-7
Potential Energy in the Field due to a Point Charge q
0
0
2
0
2
0 0
( )P
P
r
r
U r q E dl
q qk r dl
l
q qk dl
l
q q q qk k
l r
→
→
= −
= −
= −
= =
From ∞
This is also called the potential energy of
the two-charge configuration of q and q0.
What is the work required to
separate the two charges to ?
Independent of path since
static E is conservative.
Lecture 6-8
Potential Energy of a Multiple-Charge Configuration
(b)
(a)1 2 /kq q d
1 31 32 2
2
q q qq qk k k
dd
q
d++
(c)
2 3
1 3 3 41 2 2 4
1 4
2 2
q q q qq q q qk k k k
d d
q q q qk k
d
d d
d
+
+
+
+
2 d
Lecture 6-9 Electric Potential and Electric Field
1
1
1
r
q
iq
UV E dr
= = −
Normalize by the
probe charge q1:
1 int 1int
f f
q
i i
U W F d qr E dr = − = − • = −
Potential energy change of a probe charge q1 in the
electric field E created by all the other (source) charges
[Electric potential] = [energy]/[charge]
SI units: J/C = V (volts)
Potential energy difference when 1 C of charge is
moved between points of potential difference 1 V1 J =
Scalar!
independent
of q1
Lecture 6-10 Electric Potential
• So U(r)/q0 is independent of q0, allowing us to introduce
electric potential V independent of q0.
• [Electric potential] = [energy]/[charge]
SI units: J/C = V (volts)
• U(r) of a test charge q0 in electric field generated by
other source charges is proportional to q0 .
0
( )( )
U rV r
q
taking the same
reference point
Potential energy difference when 1 C of charge is
moved between points of potential difference 1 V1 J =
Scalar!
1
1
0
r
q
iq
UV E dr
= = −
Lecture 6-11
Potential at P due to a point charge q
0
0
( )( )
qU rV r
q
qk
r
=
=
From ∞
Lecture 6-12
Electric Potential Energy and Electric Potential
positive
chargeHigh U
(potential
energy)
Low U
negative
charge
High U
Low UHigh V
(potential)
Low V
Electric field direction
High V
Low V
Electric field direction
Lecture 6-13
Volt (potential) and Electron Volt (energy)
• V=U/q is measured in volts => 1 V (volt) = 1 J / 1 C
J N mV E m V
C C
N VE
C m
= = = =
= =
19
1 1 1
1 | | 1 1.602 10 1
J C V
eV e V C V−
=
• V depends on an arbitrary choice of the reference point.
• V is independent of a test charge with which to measure it.
(electron volt)
Lecture 6-15
Potential due to two (source) charges on their axis
1 2( )| | | |
q qV x k k
x x a= +
−
1 2 0q q=
Lecture 6-16
Potential due to Multiple Source Charges: Example
1 2 3 4
( )
/ 2
V P
q q q qk
d
=
+ + +
Dotted line is an equipotential when
q1=12nC, q2= -24nC, q3=31nC, q4=17nC
E from V
We can obtain the electric field E from the potential V by
inverting the integral that computes V from E:
( ) ( )
rr
x y zV r E dl E dx E dy E dz= − = − + +
If r is in x-direction, e.g., then dy=dz=0 and Ex is
negative of the rate of change of V in x-direction.
V(x)
x
0xE =