electric potential ii physics 2102 jonathan dowling physics 2102 lecture 6
TRANSCRIPT
Electric Potential IIElectric Potential II
Physics 2102
Jonathan Dowling
Physics 2102 Physics 2102 Lecture 6Lecture 6
ExampleExample
Positive and negative charges of equal magnitude Q are held in a circle of radius R.
1. What is the electric potential at the center of each circle?
• VA =
• VB =
• VC =
2. Draw an arrow representing the approximate direction of the electric field at the center of each circle.
3. Which system has the highest electric potential energy?
–Q
+Q
A
C
B€
k +3Q − 2Q( ) /r = +kQ /r
∑=i i
i
r
qkV
€
k +2Q − 4Q( ) /r = −2kQ /rk +2Q−2Q( ) / r =0
UB
Electric Potential of a Dipole (on axis)Electric Potential of a Dipole (on axis)
( ) ( )2 2
Q QV k k
a ar r
= −− +
What is V at a point at an axial distance r away from the midpoint of a dipole (on side of positive charge)?
a
r-Q +Q( ) ( )
2 2
( )( )2 2
a ar r
kQa a
r r
⎛ ⎞+ − −⎜ ⎟= ⎜ ⎟
⎜ ⎟− +⎝ ⎠
)4
(42
20
ar
Qa
−=
πε2
04 r
p
πε=
Far away, when r >> a:
V
Electric Potential on Perpendicular Electric Potential on Perpendicular Bisector of DipoleBisector of Dipole
You bring a charge of Qo = –3C from infinity to a point P on the perpendicular bisector of a dipole as shown. Is the work that you do:
a) Positive?
b) Negative?
c) Zero?
a
-Q +Q
-3C
P
U= QoV=Qo(–Q/d+Q/d)=0
d
Continuous Charge Continuous Charge DistributionsDistributions
• Divide the charge distribution into differential elements
• Write down an expression for potential from a typical element — treat as point charge
• Integrate!• Simple example: circular
rod of radius r, total charge Q; find V at center.
dqV k
r=∫
dq
r
k Qdq k
r r= =∫
Potential of Continuous Charge Potential of Continuous Charge Distribution: Example Distribution: Example
/Q Lλ = dxdq λ=
∫∫ −+==
L
xaL
dxk
r
kdqV
0)(
λ
[ ]LxaLk 0)ln( −+−= λ
⎥⎦⎤
⎢⎣⎡ +=
a
aLkV lnλ
• Uniformly charged rod• Total charge Q• Length L• What is V at position P
shown?
Px
L a
dx
Electric Field & Potential: Electric Field & Potential: A Simple Relationship! A Simple Relationship!
Notice the following:
• Point charge:– E = kQ/r2
– V = kQ/r
• Dipole (far away):– E ~ kp/r3
– V ~ kp/r2
• E is given by a DERIVATIVE of V!
• Of course!
dx
dVEx −=
Focus only on a simple case:electric field that points along +x axis but whose magnitude varies with x.
Note: • MINUS sign!• Units for E -- VOLTS/METER (V/m)
f
i
V E dsΔ =− •∫r r
Electric Field & Potential: Example Electric Field & Potential: Example • Hollow metal sphere of
radius R has a charge +q• Which of the following is
the electric potential V as a function of distance r from center of sphere?
+q
Vr1≈
rr=R
(a)
Vr1≈
rr=R
(c)
Vr1≈
rr=R
(b)
+q
Outside the sphere:• Replace by point charge!Inside the sphere:• E =0 (Gauss’ Law) • E = –dV/dr = 0 IFF V=constant
2
dVE
drd Q
kdr r
Qk
r
=−
⎡ ⎤=− ⎢ ⎥⎣ ⎦
=V
r1≈
Electric Field & Potential: Example Electric Field & Potential: Example
E2
1r
≈
Equipotentials and ConductorsEquipotentials and Conductors• Conducting surfaces are
EQUIPOTENTIALs
• At surface of conductor, E is normal to surface
• Hence, no work needed to move a charge from one point on a conductor surface to another
• Equipotentials are normal to E, so they follow the shape of the conductor near the surface.
Conductors change the field Conductors change the field around them!around them!
An uncharged conductor:
A uniform electric field:
An uncharged conductor in the initially uniform electric field:
““Sharp”conductorsSharp”conductors• Charge density is higher at conductor
surfaces that have small radius of curvature
• E = ε0 for a conductor, hence STRONGER electric fields at sharply curved surfaces!
• Used for attracting or getting rid of charge: – lightning rods– Van de Graaf -- metal brush transfers
charge from rubber belt– Mars pathfinder mission -- tungsten points
used to get rid of accumulated charge on rover (electric breakdown on Mars occurs at ~100 V/m)
(NASA)
Summary:Summary:• Electric potential: work needed to bring +1C from infinity; units = V• Electric potential uniquely defined for every point in space --
independent of path!• Electric potential is a scalar -- add contributions from individual point
charges• We calculated the electric potential produced by a single charge:
V=kq/r, and by continuous charge distributions : V=∫ kdq/r• Electric field and electric potential: E= dV/dx• Electric potential energy: work used to build the system, charge by
charge. Use W=qV for each charge.• Conductors: the charges move to make their surface equipotentials. • Charge density and electric field are higher on sharp points of
conductors.