Download January 11 - University of Utah Physics

Announcements Jan 11
• Yes the University is open all day (?)
• Webassign Online forum schedule:
– Fri, Sat, Sun, Mon. evenings 7-10pm before a
homework is due on a Tuesday morning
– See web site
• Help labs have started
– See course web page for details
http://www.physics.utah.edu/~woolf/2020_Jui/help_lab.pdf
1
18.5 Coulomb’s Law
Example A Model of the Hydrogen Atom
In the Bohr model of the hydrogen atom, the electron is in orbit
about the nuclear proton at a radius of 5.29x10-11m. Determine
the speed of the electron, assuming the orbit to be circular.
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18.5 Coulomb’s Law
Example A Model of the Hydrogen
Atom
In the Bohr model of the hydrogen atom,
the electron is in orbit about the nuclear
proton at a radius of 5.29x10-11m.
Determine the speed of the electron,
assuming the orbit to be circular.
The force on the electron is exerted by the proton, as given by Coulomb’s Law
F =k
q1 q2
r2
=
(8.99 ×10
9
)(
N ⋅ m C 1.60 ×10
2
2
(5.29 ×10
−11
m
v=
Fr
=
m
C
)
2
2
= 8.22 ×10 −8 N
mv 2
F = mac =
r
But the SAME force is responsible for the
centripetal motion
Solving for v gives
)
−19
(8.22 × 10 N )(5.29 × 10
−8
9.11 × 10-31 kg
−11
m)
= 2.18 × 106 m s
3
18.5 Coulomb’s Law
Example Three Charges on a Line
Determine the magnitude and direction of the net force on q1.
4
18.5 Coulomb’s Law
Example Three Charges on a Line
Determine the magnitude and direction of
the net force on q1.

qi q j
Fij = Fij = k
2
rij
where rij represents the distance
between qi and
F12 = k
q1 q2
F13 = k
q1 q3
r12
r13
2
2
qj
(8.99 × 10
=
(8.99 × 10
=
9
9
N ⋅ m 2 C2 )(3.0 × 10−6 C )(4.0 × 10−6 C
(0.20m )2
N ⋅ m 2 C2 )(3.0 × 10−6 C )(7.0 × 10−6 C
(0.15m )2
)
)
= 2.7 N
= 8.4 N

 
F1 = F12 + F13 = −2.7 Nˆi + 8.4 Nˆi = + 5.7Nˆi
Here ˆi denotes the unit vector in the + x direction
5
Example Three Charges not in line
Assuming the horizontal to be the x-direction,
and vertical the y-direction, determine the
magnitude and direction of the net force on q1
6
Example Three Charges not in line
Assuming the horizontal to be the x-direction,
and vertical the y-direction, determine the
magnitude and direction of the net force on q1
7
18.5 Coulomb’s Law
Example Three Charges not in line
Assuming the horizontal to be the x-direction, and
vertical the y-direction, determine the magnitude
and direction of the net force on q1
Solution: First we apply the Inverse
Square (Coulomb’s) Law to find the
magnitudes of the two forces on q1 :

qi q j
Fij = Fij = k
2
rij
where rij represents the distance
between qi and
F12 = k
q1 q2
F13 = k
q1 q3
r12
r13
2
2
qj
(8.99 × 10
=
(8.99 × 10
=
9
9
N ⋅ m 2 C2 )(4.0 × 10−6 C )(6.0 × 10−6 C
(0.15m )2
)
N ⋅ m 2 C2 )(4.0 × 10−6 C )(5.0 × 10−6 C
(0.10m )2
)
= 9.59 N
= 17.98 N
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18.5 Coulomb’s Law
The resulting force F on q1 is the vector sum of the
forces exrted by charges q2 and q3. We make this
sum component by component:
Both forces are attractive here, with dirrections
indicated in the figure to the left
Fx = + F12 cos θ 2 + F13 cos θ 3
= (9.59 N ) cos(73o ) + (17.95 N ) cos(0o )
= 2.80 N + 17.95 N = 20.75 N
Fy = + F12 sin θ 2 + F13 sin θ 3
= (9.59 N ) sin(73o ) + (17.95 N ) sin(0o )
= 9.17 N + 0 = 9.17 N
F = Fx2 + Fy2 = ( 20.75 N ) 2 + (9.17 N ) 2 = 23 N

Angle of F counter - clock - wise from from the + x - axis :
θ = tan
-1
Fy
Fx
= tan −1 (0.442) = 24o
9
18.6 The Electric Field
DEFINITION OF ELECRIC FIELD
The electric field that exists at a point is the electrostatic force experienced by a (small) test
charge q0 placed at that point divided by the charge q0 itself:

 F
E=
qo
SI Units of Electric Field: newton per coulomb (N/C)
NOTE: It is the surrounding charges that create the electric field at a given point.
The effect of the test charge itself is NEVER included in this definition
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