1. ELECTRIC FIELD ELECTRIC FLUX Lectures 3, 4 & 5 a a R 2R
In lectures 3,4 and 5, we will cover Chapter 22 which invovles calculating the electric field for different distribution of charges using 1st Coulomb’s Law and then we will introduce Gauss’ Law which is used to calculate E for symmetric shapes. a R a 2R
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2. Last Week…..Discrete Charges
Magnitude of the Electric Field due to a Point Charge: + - P
Dipole center z r+ r- q+ q- d
Electric Field due to a Dipole at point P:
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3. More on Electric Field:
Today…
More on Electric Field:
Continuous Charge Distributions
Text Reference: Chapter 22.1
Many useful examples etc.
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4. Electric Fields from Continuous Charge Distributions
Principles (Coulomb’s Law + Law of Superposition) remain the same.
Only change: + - + + + + + + + + + +
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5. Charge Distributions Problems
Step 1:Understand the geometry
Step 2:Choose dq
Step 3:Evaluate dE contribution from the infinitesimal charge element
Step 4:Exploit symmetry as appropriate
Step 5:Set up the integral
Step 6:Solve the integral
Step 7:The Result!
Step 8:Check Limiting Cases
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6. small pieces of charge dq dq =  dA =  dxdy (Cartesian coordinates)
Charge Densities
How do we represent the charge “Q” on an extended object?
small pieces of charge dq
total charge Q
Line of charge:
= charge per unit length [C/m]
dq =  dx
Surface of charge:
dq =  dA =  dxdy (Cartesian coordinates)
  = charge per unit area [C/m2]
Volume of Charge:
 = charge per unit volume [C/m3]
dq = dV =  dx dy dz
(Cartesian coordinates)
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7. Charge on a ring Steps 1-3 Problem: calculate the electric field along
z-axis due to a (circular) ring (of radius R)
of uniform positive charge (with density ).
Charge of element
The electric field due to this element
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8. Field due to charge on a ring
Step 4:Exploit symmetry as appropriate
Symmetry: direction of the field at point P must be along the positive z direction.
The z-component of the field:
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9. Field due to charge on a ring
Steps 5-7
Integrate over the entire ring with  uniform; z and R fixed
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10. Limiting cases
Step 8
z>>R: point-like charge  ~q/z2, Coulomb formula
z<<R: E=0 at the center of the ring, all field elements cancel.
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11. Quiz 1 (lecture 3)
A nonconducting semi-circular rod has a uniform charge of magnitude Q along its top half and another along its bottom half. What is the direction of the net electric field at point P? y +Q
Towards -x
Towards +x
Towards +y
Towards -y
Towards negative y. P x -Q
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12. Charged Disk
Problem: calculate the electric field along z-axis due to a (circular) disk (of radius R) of uniform positive charge (with density ).
Pick any ring element (of infinitesimal width dr) of the disk. The charge of the element is:
The electric field due to the ring element
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13. Field due to the disk Integrate over the entire disk
Use substitution of variables
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14. Field due to the disk Check Limiting Cases z>>R:
point-like charge  ~Q/z2
z>0: upper sign
z<0: lower sign
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15. Important limiting cases for charged disk
z<<R: Ez
There is a discontinuity
at z= 0. z
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16. Infinite sheet of positive chargey x
Uniform electric
fields generated on
both sides of sheet.
discontinuity in electric field at x = 0
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17. 900 arc of charge y In this coordinate systemx E
In this coordinate system
you have to deal with Ex and Ey. Y
In this coordinate system
you have to deal with the
horizontal component of E
only. E X
uniform charge distribution
total charge = Q
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18. 900 arc of charge (continued)y
rd  x dE
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19. Quiz 2 (lecture 3) (A) y (B) (C) Q x -a L (D) (E)
A thin rod with length L has a total charge of Q distributed uniformly along it’s length is located on the x-axis with one end at the origin as shown. Which expression represents the electric field due to this rod at a point on the x-axis a distance a to the left of the origin?
(A)
(B)
(C)
(D)
(E) x y -a L Q
Answer C

20. Quiz 3 (lecture 3)
Ea<Eb