Example 3-1 Determine the electric field intensity at P (-0.2, 0, -2.3) due to a point charge of 5 (nC) at Q (0.2,0.1,-2.5)

1

• Example 3-1 Determine the electric field
  intensity at P (-0.2, 0, -2.3) due to a point
  charge of 5 (nC) at Q (0.2,0.1,-2.5)
• Solution

2

• Example 3-2 Determine electric field intensity
  at an arbitrary point inside the shell. (Well do
  it by Gauss Law)

3

• Example 3-3 Velocity uniform
  electric field over a width w.
  Find the vertical deflection zL.
• Solution

4

• 3-3.1 E-Field due to Discrete Charges
• We can use the principle of superposition.

5

• The resultant force on charge q located at
  point is the vector sum of the forces exerted
  on q by each of the charges q1, q2, , qN .
• or
• The electric field intensity (electric field
  strength)
• is in the direction of and is measured
  in N/C or V/m

(force per unit charge)
6
The electric field intensity at point due to
a point charge q at . For N point charges
q1, q2, qN located at Practice exercise
Point charge 5nC is located at (2,0,4) point
charge -2nC is located at (-3,0,5) Find at
(1,-3,7)
7

• The total E-field
• Electric dipole a pair of equal and opposite
  charges q and q separated by a small distance,
  dltlt observation location, r.

P
q0
0
d
(3.23)
-q0
8

• Approximation under dltltr
• Similarly
• Hence

(3.26)
9

• Define the dipole moment
• where vector distance is from -q to q
• Then
• In spherical coordinates, by using Eq. (2-77) or

10

• 3-3.2 E-field due to distributed charges
• The differential charge rdv that contributes to
  the electric field intensity at the observation
  (field) point P is
• where
• Hence for volume distribution of charges
• For a surface distribution with charge density rs

(3.32)
(3.34)
(3.35)
11

• For a line charge
• Example 3-4 Infinitely long straight line with
  uniform charge density , find
• Solution. Because of the symmetry, we select a
  cylinder coordinate system.
• where
• Hence

(3.36)