-Electric Potential due to Continuous Charge Distributions AP Physics C Mrs. Coyle.
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Transcript of -Electric Potential due to Continuous Charge Distributions AP Physics C Mrs. Coyle.
![Page 1: -Electric Potential due to Continuous Charge Distributions AP Physics C Mrs. Coyle.](https://reader036.fdocuments.us/reader036/viewer/2022062322/5697bfc41a28abf838ca5b7c/html5/thumbnails/1.jpg)
-Electric Potential due to Continuous Charge Distributions
AP Physics C
Mrs. Coyle
![Page 2: -Electric Potential due to Continuous Charge Distributions AP Physics C Mrs. Coyle.](https://reader036.fdocuments.us/reader036/viewer/2022062322/5697bfc41a28abf838ca5b7c/html5/thumbnails/2.jpg)
Electric Potential –What we used so far! Electric Potential
Potential Difference
Potential for a point charge
Potential for multiple point charges
o
UV
q
B
Ao
UV d
q
E s
e
qV k
r
ie
i i
qV k
r
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Remember: V is a scalar quantity
Keep the signs of the charges in the equations, so V is positive for positive charges.
You need a reference V because it is changes in electric potential that are significant. When dealing with point charges and charge distributions the reference is V=0 when r
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Electric Potential Due to a Continuous Charge Distribution
How would you calculate the V at point P?
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Two Ways to Calculate Electric Potential Due to a Continuous Charge Distribution
It can be calculated in two ways: Method 1: Divide the
surface into infinitesimal elements dq
Method 2:If E is known (from Gauss’s Law) B
Ao
UV d
q
E s
e
dqV k
r
/ oE dA q
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Method 1 Consider an infinitesimal
charge element dq and treat it as a point charge
The potential at point P due to dq
e
dqdV k
r
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Method 1 Cont’d For the total potential, integrate to include the
contributions from all the dq elements
Note: reference of V = 0 is when P is an infinite distance from the charge distribution.
e
dqV k
r
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Ex 25.5 : a) V at a point on the perpendicular central axis of a Uniformly Charged Ring
Assume that the total
charge of the ring is Q.
Show that:
2 2
ee
k QdqV k
r x a
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Ex 25.5: b) Find the expression for the magnitude of the electric field at P
Start with
and
2 2
ek QV
x a
xdV
Edx
x 2 2 3/ 2Ans: E
( )
Note that at the center of the ring E=0.
How else had we calculated this result?
kQx
x a
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Ex 25.6: Find a)V and b) E at a point along the central perpendicular axis of a Uniformly Charged Disk
Assume radius a and surface charge density of σ. Assume that a disk is a series of many rings with width dr.
12 2 2Ans.: 2 eV πk σ x a x
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Ex 25.6: Find a)V and b) E at a point along the central perpendicular axis of a Uniformly Charged Disk
12 2 2Start with 2 eV πk σ x a x
2 2Ans: 2 (1 )x
xE k
x a
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Ex25.7: Find V at a point P a distance a from a Finite Line of Charge Assume the total charge of
the rod is Q, length l and a linear charge density of λ.
Hint:
2 2
Ans: lnek Q aV
a
2 2
2 2ln( )
dxx x a
x a
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Method 2 for Calculating V for a Continuous Charge Distribution:
If E is known (from Gauss’s Law)
Then use:
/ oE dA q
B
Ao
UV d
q
E s
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Ex 25.8: Find V for a Uniformly Charged Sphere (Hint: Use Gauss’s Law to find E) Assume a solid
insulating sphere of radius R and total charge Q
For r > R,
: e
QAns V k
r
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Ex 25.8: Find V for a Uniformly Charged Sphere A solid sphere of
radius R and total charge Q
For r < R,
2 23
:2
eD C
k QAns V V R r
R
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Ex 25.8:V for a Uniformly Charged Sphere, Graph The curve for inside
the sphere is parabolic
The curve for outside the sphere is a hyperbola