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