23.1 THE ELECTRIC FIELD The electric field strength E: E=F/q t (23.1) Any charged particle does not interact directly with other charged particle; rather, it responds to whatever field it encounters. E is the property of a point in space and depends only on the source of the field Q F=q E SI unit N/C (23.3) E is the resultant field strength due to all the net charges present. F=mg (g: N/kg)

Example 23.1: On a clear day there is an electric field of approximately 100N/C directed vertically down at the earth’s surface. Compare the electrical and gravitational forces on an electron.

P roblem-Solving Guide for the Electric Field 1. First draw the field vectors at the given location. (Their direction can be found by imagining that there is a positive charge at that point.) 2.Find the (scalar) magnitude of the field strength due to each charge. The signs of the charges must be ignored. This step may be ensured by writing the magnitude in the form E=k ｜ Q ｜ /r 2 (23.5) 3. Place the origin at the point at which E is being calculated. The choice of coordinate axes determines the signs of the components of the field strength E

Example 23.2: A point charge Q 1 =20 μ C is at (-d,0) while Q 2 =-10 μ C is at (+d,0). Find the resultant field strength at a point with coordinates (x,y) Take d=1.0m and x=y=2m.

The properties of lines of force 1.Electrostatics field line always start on positive charges and end on negative charges 2.The number of lines that originate from, or terminate on, a charge is proportional to the magnitude of the charge. 3.The direction of the field at a point is along the tangent to the line of force, as shown in fig. 4.The field strength is proportional to the density of the lines, that is, the number of lines per unit area, intercepted by a surface normal to the field. 5.Line of force never cross.

Example 23.3: Sketch the field lines for two point charges 2Q and –Q. (a) Symmetry (b) Near field (c) Far field (d) Null point (e) Number of lines

EXAMPLE 23.4: A proton travels a distance of 4 cm parallel to a uniform electric field E=10 3 iN/c, as shown in fig If its initial velocity is 105 m/s, find its final velocity.

EXAMPLE 23.6: A thin insulating rod of length L carries a uniformly distributed charge Q, find the field strength at a point along its axis at s distance a from one end.

EXAMPLE 23.7: What is the field strength at a distance R from an infinite line of charge with linear charge density λC/m?

EXAMPLE 23.8: nonconductoring disk of radius a has a uniform surface charge density σ C/m2. What is the field strength at a distance y form the center along the center axis?

EXAMPLE 23.9: Find the field due to the following: (a) an infinite sheet of charge with surface density +σ; (b) two parallel infinite sheets with charge densities +σ and - σ We may use the result in Ex 23.8 for the electric field due to a disk. In the limit as a→ ∞, the second term in Eq.(i) of Example 23.8 vanished and we left with E=2 π k σ, or (a) E=σ/2ε 0 (23.10) (B) E=σ/ε 0 (23.11)

Field Due to a Dipole

Torque in a Uniform Field

Potential Energy π/2

Interaction between Dipoles (Axis)

Discussion RQ : ELECTROSTATICS II Physlet physics Problem: 22.1,22.4,

Exercises of chapter 23 Questions: Exercises:25,26,35,40,44 Problem:2,4,7,11,18,19,20