1 ELECTROSTATIC

SHORT NOTES : ELECTROSTATICS.

1. COULOMB’S LAW states that the magnitude of the electrostatic (Coulomb or
electric) force between two point charges is proportional to the product of the

charges and inversely proportional to the square of the distance between them.

F = kQ1Q2 k : electrostatic Solution :cQo1n=st8a5nt1=0−96 .C0;Q12 0=950N1m0−26 C; r = 3.510−2
r2
(Coulomb)   C−2
F21 F12
Q1 Q2
F : magnitude of electrostatic force,
r : distance between two point charges, + -

 3.5 cm
where F12 : force by charge 1on charge 2
F21 : force by charge 2 on charge 1
Q1,Q2 : magnitude of charges
tohpepossaimteessigignnartetrpaecltoonneeaanano.otBhtyheaerprp–l–yirnaeFgtp1tt2uhreal=scCitvkoieQvurle1o2fQmof2robc’rsecl.eaw. equation, thus
Charges of
Charges of

- The
law
( ()( ))( )-The
sign of the charge can be ignored when sFu12b=sti9tu.0tin1g0i9n3t8.o55th11e00−−C26 o25u0lom10b−’s6
equation. coulomb (C).
S.I. unit of the charge is

- 1 Coulomb is defined as the total charge tranFs1f2e=rr3e.1d2b1y0a4 Ncurrent of one
ampere in one second.
Direction : to the left (towards Q1)

2. Q = ne e : fundamental amount of charge,1.6  10-19 C

Q : electric charge, n : positive integer number =1,2,...

3. Zero resultant electrosatatic force.

  - For two
F12 F32
Q1 Q2 Q3 dimensionan
+
+ r13 − x - x situation ( involving

r13 x –axis and y –axis)

use a resultant

FR = 0 vector concept to

F32 − F12 = 0 solve it.

 - Find FX = ?
F12 = F32
- Find FY = ?

Q1Q2 = Q2Q3 - Find FR= ?
40r122 40r232 FM = ( Fx )2 + ( Fy )2

- Find θ =?

( )1210−6 = 36 10−6 =tan −1  Fy 
x2  Fx 
3210−2 − x2

x = 0.203 m OR 20.3 cm - Find direction of θ.

4. Electric field is defined as a region of space around isolated
charge where an electric force is experienced if a positive test
charge placed in the region.

A) a single positive charge b) A single negative charge
((E Radially outward) (E Radially inward)

c) Two unequal point charges of d) Two unequal point charges of
opposite sign, +2Q and −Q opposite sign, +2Q and −Q

e) Two equal positive charges, f) Two opposite charged parallel metal
+Q and +Q f) plates.

X = neutral point

5. The electric field strength at a point is defined as the electric
(electrostatic) force per unit positive charge that acts at that

pEoin=tqFin0 ,theEsa=mkerQd2 irectiFEon::eeallseeccttthrreiiccfffoioerrlccdees.,treqn0 g:tthest charge

- It is a vector quantity.
- The unit of electric field strength is N C−1 OR V

- For two

dimensionan

situation ( involving

x –axis and y –axis)

use a resultant

vector concept to

solve it.

- Find EX = ?

- Find EY = ?

- Find ER= ?

EM = ( Ex )2 + ( Ey )2

- Find θ =?

 = tan−1 Ey 
Ex 

- Find direction of θ.

6. MOTION OF CHARGE IN ELECTRIC FIELD

Consi e a stationa y pa ti le o an is pla e in a
e e te on the ha ge is
, the
gi en by

the pa ti le has a , its
igu e . a .
, its the pa ti le has a
igu e . b .

- Consider an electron (e) with mass, me enters a uniform
electric field, E perpendicularly with an initial velocity u.

sxx v = vx2 + vy2

Zero lunch angle, θ=0˚, ++ ++ ++ +++++++ v

vx = ux = u, uy = 0 -

E sy

-u
q0 −−−− −−−−−−−−−

Figure 16.14c

a = ay = eE direction : upwards since ax = 0
me

Simulation 16.4

Zero lunch angle: The projectile Velocity x - components 50
launched horizontally. Initial velocity, u
ux = u y - components

uy = 0 ms−1

- If an object ( example the final velocity, v vx = u vy = uy − ayt
charge) is projected ax = 0 ms−2
horizontally, there should be Acceleration, a a = ay = eE = eE t
no angle of projection, so θ Sign: me me
= 0. Upward (+)
Downward (-)

Displacement, s sx = uxt sy = uyt + 1 ayt
Sign: 2
Upward (+)
Downward (-) sy = 1  eE t 2
2 me

53

- Consi e a uni o m ele t i iel is p o u e by a pai o lat
metal plates, one at whi h is ea the an the othe is at a

potential o as shown in igu e . a.

he against g aph o pai o lat metal plates an be
shown in igu e . b.

0

le t i potential o a point in an ele t i iel is e ine as

whe e

t is a quantity.

he unit o ele t i potential is .

Example 2:
Calculate electric potential at a point A and B.

Q3 + VB = VB1 + VB2 + VB3

= k Q1 + k Q2 + k Q3
B r1 r2 r3

(Put sign of the charges)

13 cm

Q1 A Q2

- 5 cm 10 cm -

VA = VA1 + VA2 + VA3

= k Q1 + k Q2 + k Q3 (Put sign of the charges)
r1 r2 r3

8. Potential difference between two points in an electr63ic field is defined as the
work done in bringing a positive test charge from a point to another point

in the electric field per unit test charge.

V = W he potential i e en e between points an B, is gi en by
q0
an

whe e

ote
the positi e test ha ge mo ing om point to point B, thus

the potential i e en e between this points is gi en by
whe e
he e o e

om the e inition o ele t i potential i e en e,
an

he e o e the hange in a potential ene gy is gi en by

Consi e a system o th ee point ha ges as shown in igu e
..

he total ele t i potential ene gy, an be e p esse as