electric field calculations for uniform ring of charge and uniformly charged disk
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Electric Field Calculations for
Uniform Ring of Charge and Uniformly Charged
DiskMontwood High School
AP Physics C
R. Casao
Electric Field of a Uniform Ring of
Charge
Montwood High School
AP Physics C
R. 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
θcosr
dqkdE:so
r
dqkdEbecomes
r
QkE
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:
3x2xr
dqxkdE
r
x
r
dqkdE
r
xθcos
3x
r
dqxkdE
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
r
Qxk :soQdq
dqr
xk
r
dqxk
21
2222
222
xaxar
xar
Combining both sides of the integration equation:
p. 652 #31, 37.
23
223
21
22
x
xa
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.
A
Qσ
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
drrσπ2dq
drrπ2σdqdAσdqdA
dqσ
dA
dq
A
Q
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:
2rQk
E
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:
2LQk
E
22 Ldrrσπ2k
dELdqk
dE
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
21
2222222
xr
xθcos
L
xθcos
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.
23
22x
21
2222x
xr
drrxσkπ2dE
xr
x
xr
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 23
22x
xr
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
23
22 u
du
2
1
u2
du
xr
drr2
dudrrdrr2du
Pull the ½ back into the equation:
21
2221
21
22
23
23
23
xr
2
u
2
21
u
22
23
uduu
u
du
21
2221
22 xr
1
xr
2
2
1
So:
21
2221
22
x
x0
1
xR
1xσkπ2xE
R
0
xσkπ2E
21
2x
2r
1
x
1
xR
1xσkπ2E
x
1
xR
1xσkπ2E
21
22x
21
221
22x
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
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