Electrostatics
principle of conservation of charge:
Pre x Conversions
You find the new charge from the previous 2 Centi (cC) x10 a
Milli (mC) x10 s
Micro (µC) x10 a
Nano (nC) x10
Pico (pC) ii
x10 a
Qnew OutQ2
2
Principle of charge quantization:
You find the number of electrons from the charge
Ne Ke
Coulomb’s law:
The magnitude of the electrostatic force exerted by one point charge (Q1) on another point
charge (Q2) is directly proportional to the product of the magnitudes of the charges and
inversely proportional to the squares of the distance between them.
k Q1 Q2 2
r2
F= force of attraction between charges Q1 and Q2 (N)
K= Coulomb’s constant (9x10gN•m2•C a)
Q= magnitudes of charges (m)
r= distance between charges (m)
Two charges experience a force F when held a distance r apart. How would this force be
affected if one charge is doubled, the other charge is tripled and the distance is halved?
F kQeQa 2 3 24F
R2 Yaya
Electrostatic force is a vector, therefore all vector rules can be applied
* direction specific
* can be added or subtracted
When calculating electrostatic force you:
◦ substitute charge magnitude only. (No signs!)
fi
, ◦Direction is determined by nature of charge (like repel, unlike attract)
◦Both objects experience the same force (Newton’s 3rd law of motion... exerts the same
force but opposites)
The net force on a straight line with 2 point charge it is co-linear
Determine the resultant electrostatic force on QB.
3cm 5cm
A B C
ane t3nc Inc
QQ't 2
10 9 r 3 1072
Fab= K(Qa)(Qb) 3 10 9 k 9 109
rs
= 9x10a(2 x 10 a)(3 x 10 a)
3x10 a
fcb= kQCQB
r
= 9x10 (1x10 a)(3 x 10 a)
(5x10 2)
FAB Fors
Fnet = Fab + Fcb s
= (-6 x 10 5) + (1,08 x 10 5)
= -4,92 x 10
Fnet= 4,92 x 10 N left
2 dimensional
Determine the resultant electrostatic force on QB.
some tame
15mm Fars
c
e Fars
9 Five
v
A
six
Fab= kQa•Qb
r
= 9 x 10 (5 x 10 a)(10 x 106)
(10 x 103)
= 4 500 N down ( A attracts B)
principle of conservation of charge:
Pre x Conversions
You find the new charge from the previous 2 Centi (cC) x10 a
Milli (mC) x10 s
Micro (µC) x10 a
Nano (nC) x10
Pico (pC) ii
x10 a
Qnew OutQ2
2
Principle of charge quantization:
You find the number of electrons from the charge
Ne Ke
Coulomb’s law:
The magnitude of the electrostatic force exerted by one point charge (Q1) on another point
charge (Q2) is directly proportional to the product of the magnitudes of the charges and
inversely proportional to the squares of the distance between them.
k Q1 Q2 2
r2
F= force of attraction between charges Q1 and Q2 (N)
K= Coulomb’s constant (9x10gN•m2•C a)
Q= magnitudes of charges (m)
r= distance between charges (m)
Two charges experience a force F when held a distance r apart. How would this force be
affected if one charge is doubled, the other charge is tripled and the distance is halved?
F kQeQa 2 3 24F
R2 Yaya
Electrostatic force is a vector, therefore all vector rules can be applied
* direction specific
* can be added or subtracted
When calculating electrostatic force you:
◦ substitute charge magnitude only. (No signs!)
fi
, ◦Direction is determined by nature of charge (like repel, unlike attract)
◦Both objects experience the same force (Newton’s 3rd law of motion... exerts the same
force but opposites)
The net force on a straight line with 2 point charge it is co-linear
Determine the resultant electrostatic force on QB.
3cm 5cm
A B C
ane t3nc Inc
QQ't 2
10 9 r 3 1072
Fab= K(Qa)(Qb) 3 10 9 k 9 109
rs
= 9x10a(2 x 10 a)(3 x 10 a)
3x10 a
fcb= kQCQB
r
= 9x10 (1x10 a)(3 x 10 a)
(5x10 2)
FAB Fors
Fnet = Fab + Fcb s
= (-6 x 10 5) + (1,08 x 10 5)
= -4,92 x 10
Fnet= 4,92 x 10 N left
2 dimensional
Determine the resultant electrostatic force on QB.
some tame
15mm Fars
c
e Fars
9 Five
v
A
six
Fab= kQa•Qb
r
= 9 x 10 (5 x 10 a)(10 x 106)
(10 x 103)
= 4 500 N down ( A attracts B)
