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Physics
Electrostatics
Charges and Coulomb’s
Law
Charges and Coulomb’s Law
positively
negatively
– Matter is made of particles which are _____________
or _____________
charged.
Coulomb
– The unit of charge is the ______________________
(C )
conserved
– Charges are ________________,
meaning that they cannot be… DESTROYED
It is thought that the total charge of the entire universe is constant and neutral.
– Charges are also ______________,
meaning that they occur in finite packages.
quantized
The smallest unit of charge is called the ______________
_____________
charge
elementary
which is equal to the charge on one proton (+) or one electron (-).
𝑞 = ±1.60 × 10−19 𝐶
Charges and Coulomb’s Law
– Coulomb determined that the force between two charged objects is
proportional to their charges and inversely proportional to the square of their
distances or:
Where:
–
𝑞1 = 1𝑠𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝐶
𝑘𝑞1 𝑞2
𝐹𝐸 =
𝑟2
𝑞2 = 2𝑛𝑑 𝑐ℎ𝑎𝑟𝑔𝑒 (𝐶)
𝑟 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑚)
𝑘 = 𝐶𝑜𝑢𝑙𝑜𝑚𝑏′ 𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 9.0 × 109
Charges and Coulomb’s Law
– There are four important things to notice from this equation.
1. Electrostatic force can be attractive OR repulsive
2. Electrostatic force depends on charge NOT mass
3. Electrostatic force gets significantly small as r increases
4. The constant k is much LARGER than the constant G
Charges and Coulomb’s Law
– There is a very important difference between gravitational and electrostatic
forces:
Gravity ALWAYS…
Attracts
Electrostatic force can… Attract or Repel
– When solving for electrostatic forces we will NOT…
Use ± signs
– Instead we will determine the direction of the force based on…
Attraction or repulsion
Example
– Two 85kg students are 1.0 m apart. What is the gravitational force between
them?
𝐺𝑚1 𝑚2
𝐹𝑔 =
M1 = 85kg
M2 = 85 kg
R = 1.0 m
𝑟2
(6.67 × 10−11 )(85)(85)
4.82 × 10−7 𝑁
𝐹𝑔 =
=
12
– If these two students each have a charge of 2.0 x 10-3 C, what is the electrostatic
force between them?
Q1 = 0.002 C
Q2 = 0.002 C
R = 1.0 m
𝑘𝑞1 𝑞2
𝐹𝐸 =
𝑟2
9 × 109 .002 (.002)
𝐹𝐸 =
= 36000 𝑁
2
1
Example
– Two point charges of 1.8 x 10-6 C and 2.4 x 10-6 C produce a force of 2.2 x 10-3 N
on each other. How far apart are these two charges?
Q1 = 1.8 x 10-6 C
Q2 = 2.4 x 10-6 C
R =?
FE = 2.2 x 10-3 N
𝑘𝑞1 𝑞2
𝐹𝐸 =
𝑟2
𝑟=
𝑘𝑞1 𝑞2
𝐹𝐸
𝑟=
(9 × 109 )(1.8 × 10−6 )(2.4 × 10−6 )
(2.2 × 10−3 )
𝑟 = 4.2𝑚
Example
FAB
A
1.7 x 10-6 C
FAC
B
2.5 x 10-6 C
C
-2.0 x 10-6 C
– A charge of 1.7 x 10-6 C is placed 2.0 x 10-2 m from a charge of 2.5 x 10-6 C and
3.5 x 10-2 m from a charge of -2.0 x 10-6 C as shown. What is the net electric
force on the 1.7 x 10-6 C charge?
𝐹𝐴𝐵
(9 × 109 )(1.7 × 10−6 )(2.5 × 10−6 )
= 95.63𝑁
=
(.02)2
𝐹𝐴𝐶
(9 × 109 )(1.7 × 10−6 )(2 × 10−6 )
= 24.98𝑁
=
(.035)2
𝐹𝑛𝑒𝑡 = 𝐵𝑖𝑔𝑔𝑒𝑟 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 = 𝐹𝐴𝐵 − 𝐹𝐴𝐶 = 95.63 − 24.98 = 70.65𝑁
Electric Field
Single Charge
Electric Field
– There are many similarities between gravitational and electrostatic forces. One
such similarity is that both forces can be exerted on objects that are not in
contact.
– In the same way that any mass is surrounded by a gravitational field, we will
imagine that any charged object is surrounded by an electric field.
– Similar to gravitational fields, an electric field will depend on:
Distance from
Size of
– ______________
and ________________
the charge.
Electric Field
– We define an electric field as the force per unit charge:
𝐹𝐸
𝐸=
𝑞
– Where:
We can substitute in Coulomb’s Law to get:
𝐸 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 (𝑁/𝐶)
𝐹𝐸 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑐𝑒 (𝑁)
𝑞 = 𝑐ℎ𝑎𝑟𝑔𝑒 (𝐶)
𝑘𝑞
𝐸= 2
𝑟
Electric Field
– In the case of electric fields we are dealing with another example of a
Force field
__________________.
vector
quantity
– Therefore the field is a ______________
_______________.
– In order to show this we always draw the field lines as
Arrows pointed in the direction a positive charge would move in the field
______________________________________________________.
– Again there is an important difference between gravitational fields and electric
fields due to the fact that… Electric fields can also repel
– We therefore define the direction of an electric field as…
+
-
Example
– What is the electric field strength at a point where a -2.00µC charge
experiences an electric force of 5.30 x 10-4 N?
Q = -2µC
FE = 5.3 x 10-4 N
E=?
𝐹𝐸 = 𝐸𝑞
(5.3 × 10−4 )
𝐸=
= 265 𝑁/𝐶
(2 × 10−6 )
Example
– At a distance of 7.5 x 10-1 m from a small charged object the electric field
strength is 2.10 x 104 N/C. At what distance from this same object would the
electric field strength be 4.20 x 104 N/C?
r1 = 0.75 m
E1 = 21000 N/C
q=?
E2 = 42000 N/C
r2 = ?
𝑘𝑞
𝐸1 = 2
𝑟1
𝐸1 𝑟12
𝑞=
𝑘
(21000)(0.75)2
= 1.31 × 10−6 𝐶
𝑞=
8.99 × 109
𝑟2 =
𝑘𝑞
=
𝐸2
((8.99 × 109 )(1.31 × 10−6 ))
= 0.53𝑚
42000
Electric Field
– We have already seen how charged particles emit electric fields, but how do
these fields interact when two or more charges act on each other?
– Consider two positively charged particles:
Electric Field
– Now, two negatively charged particles:
Electric Field
in opposition
– Note that the electric field lines point ___________________
to each other
_________________
_______________. Because this electric field is a force
add them
field, it is a vector. So when multiple fields overlap we simply _____________.
– Ok, now try two opposite charges:
connect
– Again the two fields interact, only this time they ______________________.