Montwood High SchoolMontwood High School

AP Physics CAP Physics C

R. CasaoR. Casao

Electric Field Electric Field Calculations for Line Calculations for Line of Charge Problemsof Charge Problems

Electric Field Due to a Electric Field Due to a Charged Rod on the Charged Rod on the

Axis of the RodAxis of the Rod

The Picture!The Picture!

Divide the length of the rod L into small Divide the length of the rod L into small pieces of length dx. pieces of length dx. – This also divides the total charge Q on the rod This also divides the total charge Q on the rod

into small elements of charge dq. into small elements of charge dq. – Over the length L of the rod, the charge on Over the length L of the rod, the charge on

each small piece of length dx is dq.each small piece of length dx is dq. The charge density on the rod is:The charge density on the rod is:

Because the charge is uniformly Because the charge is uniformly distributed over the length of the rod, we distributed over the length of the rod, we can set up a proportion relating the total can set up a proportion relating the total charge per unit length to the charge per charge per unit length to the charge per unit length for each small piece of the rod.unit length for each small piece of the rod.

L

Qλ

dx

dq

L

Q

Each piece of length dx contains a charge Each piece of length dx contains a charge dq and can be considered to be a point dq and can be considered to be a point charge. charge.

Each element of charge dq contributes to Each element of charge dq contributes to the net electric field E at point P.the net electric field E at point P.

Since the charge on the rod is positive, Since the charge on the rod is positive, the net electric field at point P is directed the net electric field at point P is directed along the negative x-axis from point P.along the negative x-axis from point P.

The distance from point P to each piece The distance from point P to each piece dx is x.dx is x.

Electric field equation for a point charge:Electric field equation for a point charge:

22 x

QkExr

r

QkE

The electric field contribution from each The electric field contribution from each element of charge dq is designated dE. element of charge dq is designated dE.

To determine the total electric field E at To determine the total electric field E at point P, add the electric field point P, add the electric field contribution from each element of contribution from each element of charge dq from the left end of the rod charge dq from the left end of the rod to the right end of the rod.to the right end of the rod.– Integrate from a at the left end of the rod Integrate from a at the left end of the rod

to (a + L) at the right end of the rod. to (a + L) at the right end of the rod. In other words, the electric field In other words, the electric field contributions begin at a and end at (a + L).contributions begin at a and end at (a + L).

– This means that dq needs to be expressed This means that dq needs to be expressed in terms of dx because we will integrate in terms of dx because we will integrate along the x-axis from a to (a + L).along the x-axis from a to (a + L).

For each point charge dq:For each point charge dq:

Integrate both sides of the equation.Integrate both sides of the equation.

22

22

x

dxλkdE

x

dqkdE :Substitute

dxλdqdx

dqλso

dx

dq

L

Qand

L

Qλ

x

dqkdEbecomes

x

QkE

La

a 2

La

a x

dxλkdE

The integral of 1The integral of 1·dE is E.·dE is E.

For the other side of the integration, any For the other side of the integration, any constants can come out in front of the constants can come out in front of the integral:integral:

EdELa

a

La

a

La

a

1La

a12-

12-

La

a

2-

La

a 2

La

a 2

La

a 2

x

1λkE

1

xλk

xλkE

dxxλkE

dxx

1λkE:asrewrite

x

dxλk

x

dxλkE

Substitute the (a + L) and the a into Substitute the (a + L) and the a into the -1/x part of the equation. The the -1/x part of the equation. The equation is: upper limit expression – equation is: upper limit expression – lower limit expression.lower limit expression.

Common denominator is a∙(a + L):Common denominator is a∙(a + L):

a

1

La

1λkE

a

1

La

1λkE

La

La

a

1

a

a

La

1λkE

Alternative: Alternative:

Laa

LλkE

Laa

Lλk

Laa

LaaλkE

Laa

La

Laa

aλkE

Laa

Qk

LLaa

LQkE

L

Qwithλreplaceif

Electric Field Off the Electric Field Off the Axis of a Finite Line of Axis of a Finite Line of

ChargeCharge

The Picture!The Picture!

Charge per unitCharge per unit

length:length:

L

Qλ

Divide the length of the rod L into Divide the length of the rod L into small pieces of length dx.small pieces of length dx.– This also divides the total charge Q This also divides the total charge Q

on the rod into small elements of on the rod into small elements of charge dq. charge dq.

– Over the length L of the rod, the Over the length L of the rod, the charge on each small piece of charge on each small piece of length dx is dq.length dx is dq.

The charge density on the rod is:The charge density on the rod is:

L

Qλ

Because the charge is uniformly distributed Because the charge is uniformly distributed over the length of the rod, we can set up a over the length of the rod, we can set up a proportion relating the total charge per unit proportion relating the total charge per unit length to the charge per unit length for each length to the charge per unit length for each small piece of the rod.small piece of the rod.

Each piece of length dx contains a charge dq Each piece of length dx contains a charge dq and can be considered to be a point charge. and can be considered to be a point charge.

Each element of charge dq contributes to the Each element of charge dq contributes to the net electric field E at point P.net electric field E at point P.

The total electric field E at point P is the sum The total electric field E at point P is the sum of the electric fields produced by each of the electric fields produced by each element of charge dq from –x to x.element of charge dq from –x to x.

dx

dq

L

Q

Electric field equation for a point Electric field equation for a point charge:charge:

2r

QkE

The electric field vector at point P for The electric field vector at point P for each element of charge dq can be each element of charge dq can be resolved into an x component (Eresolved into an x component (Exx)and )and a y component (Ea y component (Eyy).).

At point P, the EAt point P, the Exx components of the components of the electric field produced by each electric field produced by each symmetric –x and +x pair will be equal symmetric –x and +x pair will be equal in magnitude but opposite in direction, in magnitude but opposite in direction, therefore, these components cancel therefore, these components cancel each other.each other.

At point P, the net electric field will be At point P, the net electric field will be the sum of the Ethe sum of the Eyy components of the components of the electric field produced by each electric field produced by each element of charge dq from –x to x.element of charge dq from –x to x.

For each element of charge dq from –x to For each element of charge dq from –x to x, the values of E, r, and x, the values of E, r, and all change as x all change as x changes. The relationships between these changes. The relationships between these variables must be determined.variables must be determined.

θcosEEE

Eθcos

y

y

The electric field contribution from each The electric field contribution from each element of charge dq is designated dEelement of charge dq is designated dEyy..

To determine the total electric field E at To determine the total electric field E at point P, add the Epoint P, add the Eyy contribution from each contribution from each element of charge dq from the left end of element of charge dq from the left end of the rod to the right end of the rod.the rod to the right end of the rod.

r

yEθcosEE

r

yθcos

yx:asrewrite

yxryxr

y

21

22

22222

– Integrate from -x at the left end of the rod Integrate from -x at the left end of the rod to +x at the right end of the rod. In other to +x at the right end of the rod. In other words, the electric field contributions words, the electric field contributions begin at -x and end at +x.begin at -x and end at +x.

– This means that dq needs to be expressed This means that dq needs to be expressed in terms of dx because we will integrate in terms of dx because we will integrate from –x to +x.from –x to +x.

32y

2y2y

r

dxyλk

r

y

r

dxλkdE

dxλdqanddx

dqλso

dx

dq

L

Qand

L

Qλ

r

y

r

dqkdEbecomes

r

y

r

Qk

r

yEE

However, the radius r changes too and However, the radius r changes too and can be expressed in terms of x (in can be expressed in terms of x (in other words, r changes as x changes).other words, r changes as x changes).

Integrate both sides of the equation Integrate both sides of the equation from –x to x:from –x to x:

23

223

22y

yx

dxyλk

yx

dxyλkdE

x

x 23

22

x

xy

yx

dxyλkdE

Pull the constants k, Pull the constants k, , and y out in , and y out in front of the integral sign and integrate. front of the integral sign and integrate. The y The y22 term cannot be removed from term cannot be removed from the integration because it is being the integration because it is being added to the xadded to the x22 term. term.

x

xyy EdE

x

x 23

22y

yx

dxyλkE

From a table of integration (#98):From a table of integration (#98):

Replacing this in the equation:Replacing this in the equation:

22223

22

22223

22

yxy

x

yx

dx

aua

u

au

du

x

x222

yyxy

xyλkE

Substitute the limits of integration into the Substitute the limits of integration into the equation; remember that the equation is: equation; remember that the equation is: upper limit expression – lower limit upper limit expression – lower limit expression.expression.

x

x22

x

x222

yyxy

xλk

yxy

xyλkE

22

2222

2222

2yxy

xkE

yxy

xk

yxy

xkE

yxy

xk

yxy

xkE

y

y

y

In this example, x is equal to 0.5In this example, x is equal to 0.5··L as L as long as the center of the line of charge long as the center of the line of charge is at the origin.is at the origin.

22y

22y

yL0.25y

QkE

yL0.5Ly

L0.5Qk2E:so

LQ

λ

MIT VisualizationsMIT Visualizations

URL: URL: http://web.mit.edu/8.02t/www/802TEAL3D/visualizhttp://web.mit.edu/8.02t/www/802TEAL3D/visualizations/electrostatics/index.htmations/electrostatics/index.htm

Integrating Along a Line of Charge The Line of Charge

Electric Field Due to a Electric Field Due to a Circular Arc of ChargeCircular Arc of Charge

Consider a plastic rod having a Consider a plastic rod having a uniformly distributed charge –Q that is uniformly distributed charge –Q that is bent into a circular arc of radius r.bent into a circular arc of radius r.

The x-axis passes through the center The x-axis passes through the center of the circular arc and the point P lies of the circular arc and the point P lies at the center of curvature of the at the center of curvature of the circular arc.circular arc.

We will determine the electric field E We will determine the electric field E due to the charged rod at point P.due to the charged rod at point P.

The equation for arc length is: s = rThe equation for arc length is: s = r··.. Divide the circular arc into small, equal Divide the circular arc into small, equal

pieces of length ds.pieces of length ds.

IllustrationsIllustrations

Each length ds will contain an equal Each length ds will contain an equal amount of charge dq.amount of charge dq.

Uniform charge density allows us to Uniform charge density allows us to set up a proportional relationship set up a proportional relationship between Q, s, dq, and ds:between Q, s, dq, and ds:

s

QλsL

L

Qλ

ds

dqλ

ds

dq

s

Q

Each length ds containing charge dq Each length ds containing charge dq contributes to the net electric field at contributes to the net electric field at point P and can be considered as a point point P and can be considered as a point charge:charge:

The direction of E is towards the circular The direction of E is towards the circular arc because the charge dq is negative.arc because the charge dq is negative.

For each symmetric length ds, the Ey For each symmetric length ds, the Ey component of E are equal in magnitude component of E are equal in magnitude and opposite in direction and cancel out.and opposite in direction and cancel out.

22 r

dqkdEbecomes

r

QkE

The net electric The net electric field at point P is field at point P is the sum of the Ex the sum of the Ex components for components for each length ds each length ds from one end of from one end of the circular arc to the circular arc to the other end.the other end.

θcosdEdE

ds pieceeach for

θcosEEE

Eθcos

x

x

x

r is constant for every length ds along the r is constant for every length ds along the length of the circular arc.length of the circular arc.

Each different length ds will have a Each different length ds will have a different angle different angle between the vector E and between the vector E and the vector Ex.the vector Ex.

The dEx equation has two variables that The dEx equation has two variables that change, change, and s, therefore, we must and s, therefore, we must express one variable in terms of the other.express one variable in terms of the other.

dsr

cosθλkcosθ

r

dsλkdE

dsλdqcosθr

dqkdE

22x

2x

From the arc length equation: s = rFrom the arc length equation: s = r··– Remember that r is constant, so as s Remember that r is constant, so as s

changes so does changes so does ..

s = rs = r·· becomes d becomes ds = rs = r·d·d– dd represents the angle at point P for a represents the angle at point P for a

particular length ds.particular length ds.

dθr

θcosλkdE

dθrr

θcosλkdE

x

2x

To determine the To determine the net electric field at net electric field at point P, integrate point P, integrate from the lower end from the lower end of the rod –of the rod – to the to the upper end of the upper end of the rod +rod +..

θ

θ

θ

θdθ

r

θcosλkdEx

Left side of integral:Left side of integral:

Right side of integral: pull the Right side of integral: pull the constants k, constants k, , and r out in front of , and r out in front of the integralthe integral

x

θ

θx EdE

θ

θ

θ

θ

θ

θ

θsindθθcos

dθθcosr

λk

Combining the constants and the result of the Combining the constants and the result of the integral:integral:

Be sure to express the angle Be sure to express the angle in the correct in the correct mode on your calculator (degrees or radians).mode on your calculator (degrees or radians).

θ

θ

x

x

x

k λ k λsinθ sinθ sin θ

r rk λ

E sinθ sin θr

trig identity: sin θ sinθ

k λ k λE sinθ sinθ sinθ sinθ

r rk λ

E 2 sinθr

Keep in mind that the circular arc is Keep in mind that the circular arc is going to have a total length s that is going to have a total length s that is some part of the circumference of a some part of the circumference of a circle (C = 2circle (C = 2·· ·r). Exactly how much of ·r). Exactly how much of a circle this is will be determined by the a circle this is will be determined by the angles given.angles given.

Should the circular arc begin at 0° and Should the circular arc begin at 0° and end at 180° (or from 0 rad to end at 180° (or from 0 rad to rad), rad), substitution into the sin function will substitution into the sin function will give you an answer of 0 N/C for the give you an answer of 0 N/C for the electric field.electric field.– In that case, integrate from 0° to 90° (or 0 In that case, integrate from 0° to 90° (or 0

rad to rad to /2 rad) and multiply this by 2 since /2 rad) and multiply this by 2 since each half of the circular arc will contribute each half of the circular arc will contribute equally to the net electric field at point P.equally to the net electric field at point P.

If given a problem in which you have If given a problem in which you have a two oppositely charged circular a two oppositely charged circular arcs (one from 0arcs (one from 0°° to 180 to 180°° and the and the other from 180other from 180°° to 360 to 360°°) arranged to ) arranged to form a ring of charge, you can form a ring of charge, you can determine the electric field of one-determine the electric field of one-fourth of the circular arc and multiply fourth of the circular arc and multiply the answer by 4 since each quarter the answer by 4 since each quarter will contribute equally to the net will contribute equally to the net electric field at the point P.electric field at the point P.– Substituting the angles above into the Substituting the angles above into the

sin function results in asin function results in an answer of 0 n answer of 0 N/C for the electric field.N/C for the electric field.

Circular Arcs Within a Circular Arcs Within a QuadrantQuadrant When a circular arc lies entirely within one When a circular arc lies entirely within one

quadrant, write equations for the Equadrant, write equations for the Exx component and the Ecomponent and the Eyy component of the component of the electric field contribution for each element electric field contribution for each element of charge dq along the length of the of charge dq along the length of the circular arc.circular arc.

There is no There is no

cancellation of cancellation of

components for a components for a

circular arc withincircular arc within

the single quadrant.the single quadrant.

EExx = E = E· cos · cos

2 2

0 0

2

0

20

cos

cos

sin

sin sin02

1 0

x

x

x

x

x

x

kdE d

r

kE d

r

kE

rk

Er

kE

rk

Er

EEyy = E = E· sin · sin

2 2

0 0

2

0

20

sin

sin

cos

cos cos02

0 1

y

y

y

y

y

y

kdE d

r

kE d

r

kE

rk

Er

kE

rk

Er

Once you have the EOnce you have the Exx component and the component and the EEyy component, apply the Pythagorean component, apply the Pythagorean theorem with the two components to theorem with the two components to determine the magnitude of the resultant determine the magnitude of the resultant electric field vector.electric field vector.

Use a trig function (sin, cos, or tan) to Use a trig function (sin, cos, or tan) to determine the direction of the resultant determine the direction of the resultant electric field vector.electric field vector.