Download 1 Lesson 9 (1) Electric Potential of Uniform Surface Charge on a Cir

Lesson 9 (1) Electric Potential of Uniform Surface Charge on a Circular Disk The electric potential due to a continuous charge distribution can in principle be found by integration using the potential of a point charge. However, this is useful only in simple configurations where the integration can be performed. An example is a circular disk of radius !carrying a uniform surface charge density !, and we wish to find the potential at a point on the axis at a distance ! from the center. Divide the disk into thin rings of radius !and thickness !". The charge on the thin ring is !" = ! 2!"#" Since the point P is at the same distance ! ! + ! ! from every point on the thin ring , the potential there due to the thin ring is !"#
2!"#$%$
!" =
=
!! + !!
!! + !!
The potential at P due to the whole disk is !
!"!
! = 2!"#
!! + !!
!
With the substitution ! = ! ! , !" = 2!"!, we find !!
!"
! = !"#
= 2!"# !! + ! ! − ! ! !
!
+
!
!
Note that we take the positive sign for the square roots. When the point P is far away, we can take the limit ! → ∞, obtaining ! !
!!
!!
!"
!
! = 2!"# ! 1 + !
− ! ! ≅ 2!"# ! ! 1 + ! + ⋯ − 1 = !
2!
!
where we have used ! = !"!! for the total charge on the disk. This is the same as the potential due to a point charge. 1 In the limit of an infinite radius, which is the same as an infinite sheet charge, we get ! = ∞, which makes no sense. The reason is because we cannot choose infinity as the place where ! is zero, since there is now charge at infinity. Instead, it makes sense to choose the sheet itself as a reference where ! = 0. Then instead of using the point charge formula to find ! , we first find the electric field and integrate it to find ! . The electric field of an infinite sheet is ! = ±2!"#! where the plus sign is chosen for ! > 0and minus sign for ! < 0. Therefore, for ! > 0, !
! ! −0=−
!
!! !" = −2!"#$ and for ! < 0, the result is ! ! = 2!"#$. The two can be combined to give ! = −2!"# ! so that a plot of ! against ! when ! > 0 is as shown: (2) Electric Potential of Conductors In electrostatics, the potential at every point of a conductor is the same. The conductor is therefore an equipotential region, and the surface of the conductor an equipotential surface. To prove this, recall that the electric field is zero inside a conductor in electrostatics . Therefore, for any two points A and B inside the conductor, B !
!
VB !VA = ! # E " dr = 0 A
We can therefore speak of the potential of a conductor. Different conductors can have different potentials. (3) Electric Potential of a Spherical Conductor Consider a spherical conductor of radius a carrying a total charge Q . This charge exists only on the surface. Since the electric field outside the sphere is the same as that due a point charge Q at the center, and the electric potential is calculated from !
the integral of E , the electric potential at such a point is also the same as that due to a point charge. Thus, 2 V=
kQ
r
r ! a The potential of the conductor itself is kQ
V=
a
Example: A spherical conductor of radius R1 and charge Q1 is connected with a long metallic wire to another of radius R2 and charge Q2 . What are the charges on the conductors after the connection? Solution: Let the charges after connection be Q1! and Q2! , and the electric potentials be V1! and V2! . Since the spheres are far away, their potentials are the same as that of a single isolated sphere: kQ!
kQ!
V1! = 1 V2! = 2 R1
R2
However, since they are connected by a metallic wire, they behave as a single conductor, and so have the same electric potential: Q1! Q2!
=
R1 R2
On the other hand, from charge conservation: Q1! + Q2! = Q1 + Q2 We therefore have two equations and two unknown. The solution is R1
R
Q1! =
(Q1 + Q2 ) Q2! = 2 (Q1 + Q2 ) R1 + R2
R1 + R2
(4) Electric Potential Difference of Concentric Spherical Conducting Shells Consider a solid spherical conductor of radius a
surrounded by a concentric spherical shell of inner radius b and outer radius c . The solid conductor carries a charge Q and the conducting shell is isolated from it. To calculate the potential difference between the two conductor, we first note that from Gauss law, the electric field in the space between the conductors is radially outward and is given by 3 E=
kQ
r2
a < r < b Therefore the potential difference is b
b
#1 1&
kQ
1
Vb !Va = ! " 2 = kQ = kQ % ! ( $b a'
ra
a r
where Va is the potential of the solid sphere and Vb is the potential of the shell. The result can be written b!a
Va !Vb = kQ
ab
(5) Electric Field as Gradient of Electric Potential If equipotential surfaces are planes, V can be taken to be a function of x only. The electric field is parallel to the x-­‐axis and has x-­‐component E x only. Since the potential difference between the points x + !x and x where dx is infinitesimally small is ! !
!V = V ( x + !x ) "V ( x ) = " E # dr = "E x !x we can find E x from dV
Ex = !
dx
Thus, the electric field is equal to the negative of the slope of the potential curve plotted against x . The direction of the electric field points from the higher to the lower equipotential surface. In the general case, the equipotentials are curved surfaces, and V depends on all three rectangular coordinates x, y and z . We have shown that the electric field at any point on an equipotential surface is perpendicular to the surface. Starting from the relation ! !
!V = V ( x + !x, y + !y, z + !z ) "V ( x, y, z ) = " E # !r !
!
where !r = !xiˆ + !yĵ + !zk̂ , we can choose the infinitesimal displacement !r to be in !
!
! ! ! !
the same direction as E so that E ! "r = E "r . We then see that E points in the direction from high to low potentials, and 4 ! !V
E = ! !r
!
This relation of E with V is expressed by the equation !
E = !"V !
which says that E is the negative of the gradient of V . !
When we take !r = !xiˆ , we find !V = V ( x + !x, y, z ) "V ( x, y, z ) = "E x !x so that "V
Ex = !
"x
where in calculating the partial derivative with respect to x , the values of y and z
!
!
are held constant. By taking !r = !yĵ and !r = !zk̂ , we obtain similar relations for E y and Ez , so that !
"V
"V
"V
E = ! iˆ !
ĵ !
k̂ "x
"y
"z
Example: Determine the electric field on the axis of a uniformly charged disk. Solution: We have shown in (1) that the electric potential at a point on the z-­‐axis for z > 0 is V ( 0, 0, z ) = 2! k!
(
)
a 2 + z 2 ! z By symmetry, we expect the electric field at any point on the z-­‐axis to have only z-­‐
component. This component is given by #
&
#
&
"V
z
z
Ez = !
= !2! k" %%
!1(( = 2! k" %% !
+1(( "z
$ a2 + z2 '
$ a2 + z2 '
Example: On the x-­‐y plane, an electric dipole consists of a point charge !q at the point (!a, 0) and a point charge q at the point (a, 0) . Find the electric field at any point ( x, y ) 5 Solution: Since the distance of the point ( x, y ) from ( a, 0 ) is ( x + a)
2
( x ! a)
+ y 2 , the potential at ( x, y ) is "
%
1
1
$
' V ( x, y ) = kq
!
2
2
$
'
2
( x + a) + y 2 &
# ( x ! a) + y
Differentiating: '
+
)
)
"V
x!a
x+a
Ex = !
= !kq (!
+
, 3
2
3
2
"x
#( x + a )2 + y 2 % )
) #( x ! a )2 + y 2 %
&
$
& * $
'
+
)
)
"V
y
y
Ey = !
= !kq (!
+
, 3
2
3
2
"y
#( x + a )2 + y 2 % )
) #( x ! a )2 + y 2 %
&
$
& * $
6 2
+ y 2 and from (!a, 0 ) is