electric potential of uniform charge distributions part ii ap physics c montwood high school r....

Electric Potential of Electric Potential of Uniform Charge Uniform Charge

DistributionsDistributions Part II Part II

AP Physics CAP Physics C

Montwood High SchoolMontwood High School

R. CasaoR. Casao

Potential Difference Due to a Potential Difference Due to a Circular Arc of ChargeCircular Arc of Charge

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

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

We will determine the potential difference We will determine the potential difference V due to the charged rod at point P.V 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.


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 set up Uniform charge density allows us to set up a proportional relationship between Q, s, a proportional relationship between Q, s, dq, and ds: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 potential at contributes to the net electric potential at point P and can be considered as a point point P and can be considered as a point charge:charge:

Keep the sign of the charge in the problem Keep the sign of the charge in the problem because electric potential is a scalar because electric potential is a scalar quantity. There is no direction associated quantity. There is no direction associated with the electric potential.with the electric potential.

rdqk

dVbecomesrQk

V

Potential Difference Due to a Circular Potential Difference Due to a Circular Arc of ChargeArc of Charge

The distance of any element of charge dq to The distance of any element of charge dq to point P is a constant r (the radius).point P is a constant r (the radius).

The angle The angle with respect to the x-axis is with respect to the x-axis is different for each element of charge dq.different for each element of charge dq.

For each element of charge dq:For each element of charge dq:

dθrdsθrs

dθλkdVr

dθrλkdV

r

dsλkdVdsλdq


The net electric potential at point P is The net electric potential at point P is the sum of the contribution for each the sum of the contribution for each element of charge dq from one end element of charge dq from one end of the circular arc (of the circular arc ( = - = - /3) to the /3) to the other end (other end ( = = /3)./3).

3π

3π

3π

3π dθλkdV

Potential Difference Due to a Circular Arc of Potential Difference Due to a Circular Arc of ChargeCharge

On the left side of the equation: the sum On the left side of the equation: the sum of the contributions to the electric of the contributions to the electric potential dV at point P is the electric potential dV at point P is the electric potential V.potential V.

On the right side of the equation:On the right side of the equation:

VdV3π

3π

3π

3π

3π

3π

3π

3π

θλk

dθλkdθλk

Potential Difference Due to a Circular Arc of Potential Difference Due to a Circular Arc of ChargeCharge

Because the charge on the plastic circular arc is Because the charge on the plastic circular arc is negative, the electric potential V will be negative.negative, the electric potential V will be negative.

If the charge on the circular arc was positive, the If the charge on the circular arc was positive, the electric potential V would be positive.electric potential V would be positive.

3λkπ2

V

3π2

λk3π

3π

λkV

3π-

3π

λkθλkV 3π

3π

Electric Potential of a Uniformly Electric Potential of a Uniformly Charged DiskCharged Disk


Surface charge density:Surface charge density:

Divide the disk into concentric rings Divide the disk into concentric rings which will increase in size from the which will increase in size from the center of the disk to the outer rim of center of the disk to the outer rim of the disk.the disk.

r is the distance from the center of the r is the distance from the center of the disk to a particular ring.disk to a particular ring.

Each ring will have a different charge, Each ring will have a different charge, radius, and area. radius, and area.

A

Qσ

Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk

For each ring, as the radius changes For each ring, as the radius changes from the center of the disk to the ring from the center of the disk to the ring location, so does the amount of location, so does the amount of charge on the ring and the area of the charge on the ring and the area of the ring.ring.


For each ring:For each ring:

drrπ2dA

drr2πrdπrπddA

ringofradiusrrπA22

2

drrσπ2dq

drrπ2σdqdAσdqdA

dqσ

dA

dq

A

Q


dq is expressed in terms of dr because the dq is expressed in terms of dr because the radius of each ring will vary from the radius of each ring will vary from the center of the disk to the rim of the disk.center of the disk to the rim of the disk.

The charge within each ring can be divided The charge within each ring can be divided into equal elements of charge dq, which into equal elements of charge dq, which can then be treated as point charges can then be treated as point charges which contribute to the electric field at which contribute to the electric field at point P.point P.

R is the distance from dq to point P.R is the distance from dq to point P.


For each element of charge dq:For each element of charge dq:

R

drrσπ2kdV

R

dqkdV

R

QkVbecomes

r

QkV

212222222 xrRxrRxrR

2122 xr

drrσkπ2dV


To determine the electric potential at To determine the electric potential at point P, add each contribution from point P, add each contribution from every element of charge dq on the every element of charge dq on the disk.disk.

Integrate with respect to the radius Integrate with respect to the radius from the center of the disk (r = 0) to from the center of the disk (r = 0) to the outer rim of the disk (r = R).the outer rim of the disk (r = R).

R

0 2122

R

0xr

drrσkπ2dV


On the left side of the equation: the sum On the left side of the equation: the sum of the contributions to the electric of the contributions to the electric potential dV at point P is the electric potential dV at point P is the electric potential V.potential V.

On the right side of the equation: the 2, k, On the right side of the equation: the 2, k, , , , and x are constant and can be pulled , and x are constant and can be pulled out in front of the integral.out in front of the integral.

VdVR

0

R

0 2122 xr

drrσkπ2


This integral has to be solved by This integral has to be solved by substitution (there is no formula for this substitution (there is no formula for this integral on the integration table):integral on the integration table):– Let u = rLet u = r22 + x + x22

– Then du = 2Then du = 2·r dr + 0; du ·r dr + 0; du = 2= 2·r dr.·r dr.– The derivative of xThe derivative of x22 is 0 because it is a is 0 because it is a

constant and the derivative of a constant is 0; r constant and the derivative of a constant is 0; r is a quantity that changes.is a quantity that changes.

21

21

2122 u

du

2

1

u2

du

xr

drr2

dudrrdrr2du


The electric potential V at point P is:The electric potential V at point P is:

21

22212

1

22

21

21

21

xru221

21u

21

22

21u

21

duu21

u

du21

R0

2122 xrσkπ2V


The electric potential is:The electric potential is:

xxRσkπ2V

xxRσkπ2V

x0xRσkπ2V

2122

2122

122

21222

122

Application of Gauss’s Law Application of Gauss’s Law to Charged Insulatorsto Charged Insulators

The electric field E:The electric field E:E = 0 N/C at the middle E = 0 N/C at the middle

of the charged insulator.of the charged insulator.E increases from the E increases from the

middle of the charged middle of the charged insulator towards the insulator towards the surface.surface.

E reaches its maximum E reaches its maximum value at the surface of value at the surface of the charged insulator.the charged insulator.


To move a positive test To move a positive test charge from the middle charge from the middle of the charged insulator of the charged insulator to the outer surface, the to the outer surface, the work required increases work required increases as the test charge as the test charge moves from the middle moves from the middle of the charged insulator of the charged insulator to the outer surface.to the outer surface.


The maximum amount The maximum amount of work is performed of work is performed as the test charge as the test charge reaches the outer reaches the outer surface.surface.

No additional work is No additional work is required to move the required to move the test charge beyond test charge beyond the outer surface.the outer surface.


Inside the sphere:Inside the sphere:Low voltage occurs in Low voltage occurs in

center of the sphere.center of the sphere.High voltage occurs at the High voltage occurs at the

outer surface of the sphere.outer surface of the sphere.Electric field lines inside the Electric field lines inside the

uniformly charged sphere uniformly charged sphere are directed radially inward are directed radially inward from the outer high voltage from the outer high voltage surface to the low voltage surface to the low voltage center. center.


Electric field lines extend Electric field lines extend radially outward from the radially outward from the high voltage surface to 0 J/C high voltage surface to 0 J/C at at ..

From From to the surface, to the surface, the work needed to move the work needed to move a positive test charge a positive test charge from from to the surface to the surface increases from 0 J to a increases from 0 J to a maximum value at the maximum value at the surface of the sphere.surface of the sphere.

electric potential of uniform charge distributions part ii ap physics c montwood high school r....

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