electric field ring of charge & disk
TRANSCRIPT
Electric Field Calculations for
Uniform Ring of Charge and Uniformly Charged
DiskMontwood High SchoolAP Physics C
R. Casao
Electric Field of a Uniform Ring of
ChargeMontwood High School
AP Physics CR. Casao
Consider the ring as a line of charge that has been formed into a ring. Divide the ring into equal elements of charge dq; each element of charge dq is the same distance r from point P. Each element of charge dq can be
considered as a point charge which
contributes to the net electric field at point P.
At point P, the electric field contribution from each element of charge dq can be resolved into an x component (Ex) and a y component (Ey).
The Ey component for the electric field from an element of charge dq on one side of the ring is equal in magnitude but opposite in direction to the Ey component for the electric field produced by the element of charge dq on the opposite side of the ring (180º away). These Ey components cancel each other.
The net electric field E lies completely along the x-axis.
Each element of charge dq can be considered as a point charge:
θcosEEE
Eθcos x
x
θcosrdqkdE:so
rdqkdEbecomes
rQkE
2x
22
cos can be expressed in terms of x and r:
The total electric field can be found by adding the x-components of the electric field produced by each element of charge dq.
Integrate around the circumference of the ring:
3x2x rdqxkdE
rx
rdqkdE
rxθcos
3x rdqxkdE
is the symbol for integrating around a closed surface. Left side of the integral: adding up all the
little pieces of dEx around the circumference gives us Ex (the total electric field at the point).
Right side of the integral: pull the constants k, x, and r out in front of the integral sign.
xx EdE
However, r can be expressed in terms of the radius of the ring, a, and the position on the x-axis, x.
3
33
rQxk :soQdq
dqr
xkr
dqxk
212222
222
xaxar
xar
Combining both sides of the integration equation:
p. 652 #31, 37.
23223
2122
xxa
Qxk
xa
QxkE
MIT Visualizations
URL: http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/electrostatics/index.htm
The Charged Ring Integrating Around a Ring of Charge
Electric Field of a Uniformly Charged
Disk
Surface charge density:
Divide the disk into concentric rings which will increase in size from the center of the disk to the outer rim of the disk.
r is the distance from the center of the disk to a particular ring.
Each ring will have a different charge, radius, and area.
AQσ
For each ring, as the radius changes from the center of the disk to the ring location, so does the amount of charge on the ring and the area of the ring.
For each ring:
drrπ2dA
drr2πrdπrπddA
ringofradiusrrπA22
2
drrσπ2dqdrrπ2σdqdAσdq
dAdqσ
dAdq
AQ
dq is expressed in terms of dr because the radius of each ring will vary from the center of the disk to the rim of the disk.
The charge within each ring can be divided into equal elements of charge dq, which can then be treated as point charges which contribute to the electric field at point P (see the ring problem).
Point charge equation:
2rQkE
The distance from the point charge to the point P (r) was labeled as L in the picture.
The contribution of each element of charge dq to the net electric field at point P is:
2LQkE
22 Ldrrσπ2kdEL
dqkdE
At point P:
The y-components for each opposite charge dq cancels; only the x-components contribute to the net electric field at point P.
This is true for every ring. The net electric field is given by:
Substitute:
θcosEEE
Eθcos x
x
θcosL
drrσkπ2dE 2x
Express the cos in terms of the variables x and r. L is the distance from dq to point P.
22
212222222
xr
xθcosLxθcos
xrLxrLxrL
Integrate with respect to the radius from the center of the disk (r = 0) to the outer rim of the disk (r = R).
The 2, k, , , and x are constant and can be pulled out in front of the integral.
2322
x
212222x
xr
drrxσkπ2dE
xr
xxr
drrσkπ2dE
Left side of the equation: adding all the x-components together gives us the net electric field, Ex.
Right side of the equation: this integral has to be solved by substitution (there is no formula for this integral on the integration table):
R
0
R
0 2322
xxr
drrxσkπ2dE
x
R
0x EdE
Substitution method: Let u = r2 + x2
Then du = 2·r dr + 0; du = 2·r dr. The derivative of x2 is 0 because it is a
constant and the derivative of a constant is 0; r is a quantity that changes.
23
23
2322 u
du21
u2
du
xr
drr2
dudrrdrr2du
Pull the ½ back into the equation: 2
12221
21
22
23
23
23
xr
2
u
2
21
u
22
23
uduuu
du
21222
122 xr
1
xr
221
So:
21222
122
x
x0
1
xR
1xσkπ2xE
R
0
xσkπ2E2
12x2r
1
x1
xR
1xσkπ2E
x
1
xR
1xσkπ2E
2122
x
2122
122x
For problems in which x is very small in comparison to the radius of the disk (x << R), called a near-field approximation:
p. 652, #33, 34, 37 (2nd half) Homework 3, #5
σkπ2E x