24: Chapter Electric Potential Part B
! ! ! ! ! ! !
Electric potential energy and potential. Calculating the potential from the field. Potential due to a point charge. Potential due to a group of point charges. Potential due to an electric dipole. HITT question. Potential due to a continuous charge distribution Line of charge. Charged disk.
Ch. 24: Electric Potential - Part B 1
September September 16, 2009
Electric Potential Energy
initial state
q1 q2
final state
q1 q2 q3
q3
Ui
Uf
If a system of charges changes its configuration from an initial state i to a different final state f, the electrostatic force does work W on the particles. The work done is path independent (conservative force).
U = U f Ui = W
September September 16, 2009 Ch. 24: Electric Potential - Part B 2
Electric Potential
!
The electric potential V is the potential energy per unit charge at a point in an electric field.
U V= q
! !
The electric potential is a scalar (not a vector). Electric potential difference between two points.
U V = Vf Vi = q !! !! U = W = F d = qE d !! V = E d
September September 16, 2009 Ch. 24: Electric Potential - Part B 3
Calculating the Potential from the Field
!
We can calculate the potential difference between any two points i and f in an electric field if we know the electric field vector E all along any path connecting those points.
!
!
First we calculate the work done on a positive test charge +q0 by an arbitrary (nonuniform) electric field E as the charge moves along the path from i to f. At any point on the path, an electrostatic force q0E acts on the charge as it moves through a differential displacement ds.
!! ! ! dW = F ds for F = q0 E
September September 16, 2009
!! dW = q0 E ds
Ch. 24: Electric Potential - Part B
4
Calculating the Potential from the Field
!
We find the total work by summing (integrating) the differential works as the charge moves through all the displacements ds along the path.
!! W = dW = q0 E ds
i
f
W V = Vf Vi = q0 f !! Vf Vi = E ds
The potential difference between any two points i and f in an electric field is the negative of the!line!integral (integral along a particular path) of E ds from i to f.
September 16, 2009 Ch. 24: Electric Potential - Part B 5
i
Calculating the Potential from the Field
!
If we set Vi=0, then
!! V = E ds
!
f
The above equation also gives the potential at any point f relative to the zero potential at infinity.
i
September 16, 2009
Ch. 24: Electric Potential - Part B
6
Calculating the Potential from the Field
Charge q0 is moved between points i and f in an uniform electric field E. Points i and f lie on the same electric field line. Calculate the potential difference Vf-Vi for the path icf. Verify that you will get the same result for the direct path if!
!! ! ! Vc Vi = E ds = 0 since E ds , and =90
c if
f !! Vf Vc = E ds = E cos( 90 ) ds = E (sin ) dcf
d = E (sin ) = Ed sin
September 16, 2009
c
c
Ch. 24: Electric Potential - Part B
7
Potential due to a Point Charge
!
!
!
!! V = E ds
i
f
!
The positive point charge q produces an electric field E and an electric potential V at point P (at distance R from the point charge). We move a test charge q0 (shown at a distance r from the point charge) from point P to infinity, to find the potential. The electric field E is directed radially outward from the fixed charge q. For the integration we take the direct path, a line that extends radially from the fixed charge through point P to infinity.
Ch. 24: Electric Potential - Part B 8
September 16, 2009
Potential due to a Point Charge
!
! The differential displacement ds of the test charge along its path has the same ! direction as E . !! E ds = E cos ds = E cos 0 ds = Eds
!! Vf Vi = E ds = Edr i R
ds = dr
since the path is radial
f
Vf = V ( ) = 0 and Vi = V ( R ) = V
0 V = Edr where E =
R
September 16, 2009
1
q
4 0 r 2
for the electric field of a point charge
9
Ch. 24: Electric Potential - Part B
Potential due to a Point Charge
0 V =
1
q
4 0 r 2 R
q
dr =
q
4 0
r
R
1
2
dr
q 1 = r = 4 4 0 R 0
q 1 0 = 4 0 R R
V=
1
q
4 0 r
general result, r is the radial distance from the point charge, q can be any point charge (+,-,0)
A positively charged particle produces a positive electric potential, while a negatively charged particle produces a negative electric potential.
September 16, 2009 Ch. 24: Electric Potential - Part B 10
Potential due to a Point Charge
A computer-generated model of V(r) for points on the (x,y) plane. The (positive) charge is at (0,0).
V (r) =
1
q
4 0 r
The same equation is valid for the electric potential outside or on the external surface of a symmetric spherically charge distribution (using the shell theorem for the electric field).
V=
1
q
4 0 r
q is the net charge of the sphere or shell, and r is the radial distance from the center of the charge distribution
Ch. 24: Electric Potential - Part B 11
September 16, 2009
Potential due to a Group of Point Charges
!
The net electric potential V for a group of point charges is found by applying the superposition principle.
V = Vi =
1
!
n
1 4 0
r
1
n
qi
i
!
This is an algebraic sum, the signs of the charges are important. Much easier to calculate than a vector sum, such as when calculating the net electric field.
September 16, 2009
Ch. 24: Electric Potential - Part B
12
Potential due to a Group of Point Charges
!
Examples. The charges are two protons. Rank the potentials for the three arrangements.
q1 q2 q 1 1 V = V1 + V2 = + = + r2 4 0 d D 4 0 r1 1
The electric potential is a scalar and does not depend on the orientation. The three potentials are equal!
September 16, 2009 Ch. 24: Electric Potential - Part B 13
Potential due to a Group of Point Charges
!
2Q r
The charges are located at the corners of a square of side d=0.24 m. Calculate the potential at point P at the center of the square. 4 4 4Q 1 4 qi 1
V = Vi =
1
d
P
q
1
4
4 0
1
r
=
4 0
q r
1
i
i
= q1 + q2 + q3 + q4
= 2Q + 4Q +Q Q = 6Q
-Q
+Q
r = d / 2 and if Q=1.5 nC, then
6 Q 2 ( 8.99 109 Nm2 /C2 )( 6 1.5 109 C ) 2 V= = = 477 V 4 0 d 0.24 m
September 16, 2009 Ch. 24: Electric Potential - Part B 14
Potential due to an Electric Dipole
!
!
!
An electric dipole with charges +q and q at a distance d apart. Calculate the potential at an arbitrary point P (at a distance r from the center of the dipole). Let r(+) and r(-) are the distances from P to the +q and q charges, respectively; and V(+) and V(-) are the potentials produced by the +q and q charges at point P, respectively.
V = V( + ) + V( )
q q = + 4 0 r( + ) r( ) 1
q r( ) r( + ) = 4 r r 0 () (+)
15
September 16, 2009
Ch. 24: Electric Potential - Part B
Potential due to an Electric Dipole
!
For distances r much further than the size d of the dipole, we can use the following approximations.
r( ) r( + ) d cos r( ) r( + ) r 2
! p
V= V=
q 4 0 1 4 0
r( ) r( + ) r( ) r( + ) p cos r2
q 4 0
d cos r2
where p=qd is the magnitude of the electric dipole moment.
September 16, 2009 Ch. 24: Electric Potential - Part B 16
HITT Question
How many hours per week do you spend studying for PHY2049? A) 0-2 hours; B) 2-4 hours; C) 4-6 hours; D) 6-8 hours; E) >8 hours.
September September 16, 2009
Ch. 24: Electric Potential - Part B
17
Electric Potential
!
Potential due to a continuous charge distribution.
!
dq
For continuous charge distribution, we can not use summation to find the electric potential, but must use integration. ! We choose a differential element of charge dq, determine the potential dV at point P due to dq (treated as a point charge), and integrate over the entire distribution.
dV =
1
dq
4 0 r
then
V = dV =
1 4 0
dq r
18
September 16, 2009
Ch. 24: Electric Potential - Part B
Electric Potential due to a Line of Charge
!
dq
L
A thin nonconducting rod of length L has a positive charge of uniform linear density . Calculate the electric potential V due to the rod at point P, a perpendicular distance from the left end of the rod.
dq = dx
where = q / L
r = d2 + x2 dV = 1 dq 4 0 r = 1
dx
4 0 x 2 + d 2
(
)
1/2
September 16, 2009
Ch. 24: Electric Potential - Part B
19
Electric Potential due to a Line of Charge
V = dV =
0
L
1 4 0 dx
x2 + d2
dx
V= 4 0
dq
0
L
L
V= 4 0 V= 4 0
l n L + x 2 + d 2 0
( l n L + (
x2 + d2
L2 + d 2
) ln d )
L
20
V>0 since q>0.
V= 4 0
L + L2 + d 2 l n d
September 16, 2009
Ch. 24: Electric Potential - Part B
Electric Potential of a Charged Disk
!
!
!
A plastic disk of radius R, charged on its top surface to a uniform surface charge density . Find the potential V at point P on the central axis of the disk at a distance z from the center. We choose a differential element consisting of a flat ring or radius R and radial width dR. Its charge has a magnitude
dq = dA = d R ' 2 = ( 2 R ' )( dR ' )
!
(
)
All points of the charged element are the same distance r from point P.
r = z 2 + R '2
September 16, 2009 Ch. 24: Electric Potential - Part B 21
Electric Potential of a Charged Disk
dV =
1
dq
4 0 r
=
1 4 0
( 2 R ' )( dR ' )
z 2 + R '2
V = dV = 2 0
0
R
R ' ( dR ' ) z 2 + R '2
z>0
R 2 V= z + R '2 0 2 0
dy y
=2 y
V= 2 0
(
z 2 + R2 z
)
y = R '2 + z 2
September 16, 2009
Ch. 24: Electric Potential - Part B
22