Electrostatics

Coulomb’s Law

Coulomb’s Law

F = (kq1q2)/r2

K = 9 x 109 Nm2/C2

Positive force means repulsionNegative force means attraction

Example problem

A charge q1 = 4 μC is positioned at the origin. A charge of q2 = 9 μC is positioned at x = 4 m. Where on the x-axis can a charge q3 be placed so that the force is zero?

q1 q3 q2

4 m

Solution

(kq1q3)/x2 = (kq2q3)/ (4-x)2

kq1(4-x)2 = kq2 x2

4 μC (4-x)2 = 9 x2

4(16-8x+x2) = 9x2

5x2 +32x – 64 = 0 x = 1.6 m

Example problem

Table salt is a crystal with a simple cubic structure with Na+ and Cl- ions alternating on adjacent lattice sites. The distance between ions is a = 2.82 x 10-

10 m. a)What force does Na+ experience due to one of its Cl- neighbors? B)What force does a Cl- ion experience due to a neighboring Na+? c)What force does an Na+ ion at the origin experience due to Cl- ions at (a,0,0) and (0,a,0)?

Solution

A) F = kq1q2/r2 = [9 x 109 (1.6 x10-19

C)2]/(.282 x 10-10 m)2 = 2.9 x 10-9 NB) same as a by Newton’s Third LawC) F = F1 + F2 = 2.9 x 10-9 (i + j) N = √(2.9 x 10-9)2 + (2.9 x 10-9 )2 = 4.1 x 10-9 N

Electric Field

Vector QuantityAt every point in space it has a magnitude and directionThe total electric field at any point is the sum of the electric fields due to all charges that are presentUnit: N/CAlways point away from positive charge and toward negative charge

Problem

Find the force on a Ca+2 ion placed in an electric field of 800 N/C directed along the positive z-axis.

Solution:F = qEq = 2eF = 2e(800N/C)2.56 x 10-16 N

Electric Field

E = F/qF = kqQ/r2

E = kqQ/qr2

E = kQ/r2

Problem 2

A point charge q = -8.0nC is located at the origin. Find the electric field vector at the field point x = 1.2 m and y = -1.6 m.

Solution

E = - 11 N/C i and 14 N/C j

Problem 3

Four identical charges are placed on the corners of a square of side L. Determine the magnitude and direction of the electric field due to them at the midpoint of one of the square’s sides.

Solution…

P

A

C

B

D

EA

EC

For B and D the electric fields are the same and opposite so thy cancel each other out.

E = 2kq/(cos2θL2)

Problem 4

An electric dipole consists of +q and –q and separated by a distance of 2a. If the charges are positioned at (0,0,a and (0,0,-a) determine the electric field at a point a distance of z from the origin on the z-axis, where z >> 2a.

Charge Distributions and E-fields

ΔE = k ΔQ/r2 rE = k Σi Δqi/ri

2 r

E = k lim Σ Δqi/ri2 r = k ∫dq/r2 r

Linear charge distribution

Charge Q is distributed on a line of length l, the linear charge densityλ = Q/ldq = λ dl

Surface Area Charge Distribution

Charge Q is distributed on a surface area A, the surface densityδ = Q/Adq = δ dl

Volume Charge Distribution

Charge Q is distributed uniformly throughout a volume V, the volume charge densityρ = Q/Vdq = ρ dl

Problem 1

A rod of length l has a uniform positive charge per unit length, λ, and a total charge, Q. Calculate the electric field at a point P along the axis of the rod and a distance a from one end.

Problem 2

A ring of radius a carries a uniformly distributed positive charge Q. Calculate the E field due to the ring at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring.