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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.
Potential Difference Due to a Potential Difference Due to a Circular Arc of ChargeCircular Arc of Charge
Potential Difference Due to a Potential Difference Due to a Circular Arc of ChargeCircular Arc of Charge
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
Potential Difference Due to a Potential Difference Due to a Circular Arc of ChargeCircular Arc of Charge
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
Potential Difference Due to a Potential Difference Due to a Circular Arc of ChargeCircular Arc of Charge
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
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 Electric Potential of a Uniformly Charged DiskCharged Disk
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.
Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk
For each ring:For each ring:
drrπ2dA
drr2πrdπrπddA
ringofradiusrrπA22
2
drrσπ2dq
drrπ2σdqdAσdqdA
dqσ
dA
dq
A
Q
Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk
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.
Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk
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
Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk
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
Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk
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
Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk
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
Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk
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
Electric Potential of a Uniformly Charged Electric Potential of a Uniformly Charged DiskDisk
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.
Application of Gauss’s Law Application of Gauss’s Law to Charged Insulatorsto Charged Insulators
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.
Application of Gauss’s Law Application of Gauss’s Law to Charged Insulatorsto Charged Insulators
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.
Application of Gauss’s Law Application of Gauss’s Law to Charged Insulatorsto Charged Insulators
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.
Application of Gauss’s Law Application of Gauss’s Law to Charged Insulatorsto Charged Insulators
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.
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