EEE340Lecture 081 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

EEE340Lecture 082 Example 3-2 Determine electric field intensity at an arbitrary point inside the shell. (We’ll do it by Gauss’ Law)

EEE340Lecture 083 Example 3-3 Velocity uniform electric field over a width w. Find the vertical deflection z=L. Solution

EEE340Lecture E-Field due to Discrete Charges We can use the principle of superposition.

EEE340Lecture 085 The resultant force on charge q located at point is the vector sum of the forces exerted on q by each of the charges q 1, q 2, …, q N. 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)

EEE340Lecture 086 The electric field intensity at point due to a point charge q at. For N point charges q 1, q 2, … q N 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)

EEE340Lecture 087 The total E-field Electric dipole: a pair of equal and opposite charges +q and –q separated by a small distance, d<< observation location, r. (3.23) +q0 -q0 0d P

EEE340Lecture 088 Approximation under d<<r Similarly Hence (3.26)

EEE340Lecture 089 Define the dipole moment where vector distance is from -q to +q Then In spherical coordinates, by using Eq. (2-77) or

EEE340Lecture : E-field due to distributed charges The differential charge  dv’ 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  s (3.34) (3.35) (3.32)

EEE340Lecture 0811 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)