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Electrostatics
Coulomb’s Law
Coulomb’s Law
F = (kq1q2)/r2
K = 9 x 109 Nm2/C2
Positive force means repulsion
Negative 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 xaxis can a charge q3
be placed so that
the force is zero?
4m
q1
q3
q2
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 Clneighbors? 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 N
B) same as a by Newton’s Third Law
C) 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 Quantity
At every point in space it has a magnitude
and direction
The total electric field at any point is the sum
of the electric fields due to all charges that
are present
Unit: N/C
Always 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 zaxis.
Solution:
F = qE
q = 2e
F = 2e(800N/C)
2.56 x 10-16 N
Electric Field
E = F/q
F = 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…
B
A
EC
For B and D the
electric fields are
the same and
opposite so thy
cancel each other
out.
P
EA
C
D
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 Efields
ΔE = k ΔQ/r2 r
E = k Σi Δqi/ri2 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 density
λ = Q/l
dq = λ dl
Surface Area Charge
Distribution
Charge Q is distributed on a surface
area A, the surface density
δ = Q/A
dq = δ dl
Volume Charge Distribution
Charge Q is distributed uniformly
throughout a volume V, the volume
charge density
ρ = Q/V
dq = 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.