Charge Coulomb‛s Law Equillibrium of Charges Charge on pendulum Electric Field Neutral Point Charged particle released
in an electric field
PHYSICS
WALLAH Quantization of charge
Q1 F F Q2 F=
1
4πε0
Q 1Q 2 E Electric field at a point, due to
Like Charges
Q=+
r2
qE
point charge Kq
1) Force, F=qE
-ne Q=Total charge E =
r Calculation of Charge tanθ= mg 1 r2
ε0=Permitivity of free space E
r2
Q1 r qE
n=1,2,3.... l
(
(
r1 r2
θ θ E
K=
1 x1 = 2) Acceleration, a= m
e=1.6 x 10-19C q q 4πε0 Q1 + Q 2
Q1 Q2 Sin θ= r
r
q
r
qE x1 x2 qE
2l 3) Velocity, V= t
2
< < N
Superposition Q1 Q2
m
Additivity of charge ε0 = [Q
[ ] 1
] [Q2]
[r2] [F]
=
[AT] [AT]
[L2] [MLT-2]
= M-1L-3T4A2 r = 2l sin θ mg r Q2 r
Q1 r1 2 E1
E x-
2= 4) Velocity, V= 2qE
Ql =Q1+Q2
( (
General rule
> net
x
>
Fair = q in equillibrium Q1 + Q2
r2 if θ is very small
F
Fmed= k Q2 ENet = m
Redistribution of charge Q1
F
k Q2 tanθ Sin θ θ
> E2 q2E2t2
Q2
r 5) Kinetic energy, K.E =
Q1
Unlike Charges 2m
r r k=dielectric constant of the medium r = qE E1
r1 2 2l mg ENet = E1 + E2
Q l=
Q1+Q2
Superposition q =-
( r1+r2 ( Q2 Q 1 in equillibrium
r /
k q2 r2
r3 ∝ q2
E2 Outside closer to smaller charge E E
-
+
2 Direction: = ,
E1
ENet = E1 - E2 accelerated in the accelerated opposite
Ql=Charge on each shell after redistribution a) Like- Towards the point at which force has to
2l mg Q1 Q2
E2
N direction of electric to the direction of
be evaluated (repulsion) field electric field
Charge Density b) Unlike- Away from the point at which force E1 r
r2 2 E
> net ENet =
(
Vy
( E12 + E22 +E1E2
>
Linear Charge density, λ=Q
C has to be evaluated (attraction) q =- Q1 Q 2 in equillibrium <|Q |
ELECTROSTATICS - 1
Unit= m r1+r2 If, E1=E2=E Then, Enet= 3 E 1 V = Vx 2+Vy 2
L F1
F
E V
General rule 60o
> E2
Q > net
>
q
Surface Charge density, σ =
( (
C
= U2+ qE t
2
Unit=
S m2 Vx = qEt
Density of ball “ρ” Q2 r m
(
u
(
x=
θ θ M
Q
Volume Charge density, ρ = Unit=
C θ
> F2 Enet
Q1 - Q2
>
V m3 accelerated in the direction of field and
>
F
F E1 ENet = E12 + E22 perpendicular to initial velocity
Q=Total charge V=Volume > net When θ =60o Distance from Q1=x+r
>
L=Length S=Area it θ does not change on submerging in liquid > If, E1=E2=E Then, Enet= 2 E
Fnet= 3F E2
MP
Dielectric constant of liquid, = 1837 , e = 1.7 10 -11
60o
+
>F m
Enet ENet = E12 + E22 -E1E2 Two point charges +8q and -2q are located Me 1 at2=h=Constant
E1 at x = 0 and x = L respectively.The location of a point
Fnet When θ =90o A charge is placed at the centre of 2
>
If, E1=E2=E Then, Enet= E
120O
>
on the x axis at which the net electric field due to E
p e
If a charge on the body is 1 nC,then 1 qE 2
the line joining two equal charges Q. The system ρ E2
(
(
θ θ
these two point charges is zero is:
how many electrons are present on the body? Fnet= 2 F t = h
2 m
F K= E
of the three charges will be in equilibrium if q ρ- Direction h
is equal to a) 8 L t1 h t2
a) 1.6 × 1019 b) 6.25 × 109 F
> t2 m
1) Positive charge:-Towards the point at which
tp m 1/2
[ [
b) 4 L
c) 6.25 × 1027 d) 6.25 × 1028 Fnet
density of liquid= electric field has to be evaluated = p
When θ =120o
a) -Q/2 c) +Q/4 c) 2 L te me
F Fnet= F 2) Negative charge:-Away from the point at which
b) -Q/4 d) +Q/2 electric field has to be evaluated d) L/4 tp>te
120O
F
3) Charge placed at the face
Time period of Charged Properties of field lines Electric flux Application of Gauss‛s Theorem 9) Electric field due to circular arc at its
center
Pendulum in an electric filed Flux is proportional to λ
Kq
1) Start from positive charge and end on total no. of field lines
passing through an area
q 1) Point charge E=
r2 E
++
++++ +++ ++
+++
Φ =
++ ++
++
negative charge s E
20
++
cube
QE l ∫
Φ = E.ds cos θ Qouter Qinner
T=2 π Φ =∫ E.ds Qinner= Qplate- Qouter
m)
(g- QE 2) Never intersect each other. If they intersect r
there will be 2 directions for electric field
2) Metal sphere/Hollow sphere
Eo= 2k
r
λ sin(θ/ )
2
mg which is not possible r
Gauss Law:- Φ = q0 = ∫0 E.ds cos θ
KQ
Tangent
E Time period will increase + +
P
4) Charge placed at the corner Esurface= +
electric lines
of force
Tangent
R2 +
R
+ E
+ Qplate θ
qnet KQ +
+
eg: For a semicircle θ=180o
E Zero flux:- Φ = = 0,where qnet=0 Φcube=
Q Eoutside= 2
r
r + +
+
R r
O
O
8 0
Eo= 2k λ sin(180/ )
0
l 3) P
o
180o
Always perpendicular to Q 1 Q r 2
T=2 π
++++
+++
Conducting surface Φone face= =
m)
(g+ QE Einside= 0
8 0 3 24
= 2k λ
++
QE Electric flux for Cube
+
++
0
++
r ++
++
++
mg 7) Electric field due to a finite linear ++
+++
++++++
++
1) No charge inside the cube 3) Non-Conducting sphere charge distribution
Time period will decrease 10) Electric field at the center of a
KQr circular ring
E α Electric field
+
4) Charge inside q=0 Einside= +
+
+++ ++++
E line density R3 + ++
++
q
+++ +++
+
l KQ E
+++ +++
QE 5) Charge placed at the edge O
T=2π E1
Φ= =0 Esurface= 2 λ +
Eo=0
++
R 2k λ +
++
r
(g2+( QEm )
2
E= r sin(θ/2) + (
mg
E1 > E2 0 θ
Time period will decrease KQ +
++
++
Eoutside= R r + +++ +++
E2
Charge inside q=0
Q r2 +
++
Φcube= +
4 0 11) Electric field due to a circular ring
Electric field inside a 2) Charge placed at the center Q Q +
Φface= 0 1 = 16 4) Conducting sheet of charge
dielectric medium 4
+
5)
0
Never form closed loops (Conservative force)
8) Electric field due to a infinite linear
q charge distribution E= π
4 ε
1 qx x2+r2
6) q α no. of field lines
Φ = 0
(x2+r 2)3/2
r
| q2| > | q1|
total
0 E= ∞
(For large distance)
0 2k λ θ
E= r x
Enet= E K q1 q2 6) Flux through curved surface
+
E= π
1 q
k +
+
4 ε0 x2
+
5) Non-conducting sheet
q
+
Electric lines of force about negative + r
λ
=
P
point charge are: Φ + E
a) circular, anticlockwise
one side
6 0 +
+ Emax
Φeffective = Φcurve+ 2Φcross section + + + + +
Enet=E-Einduced, Einduced=E-Enet =E (1- 1K )
b) circular, clockwise E= +
c) radial, inward 2 0
+ + + + +
x
x=r
d) radial, outward 2
in an electric field
PHYSICS
WALLAH Quantization of charge
Q1 F F Q2 F=
1
4πε0
Q 1Q 2 E Electric field at a point, due to
Like Charges
Q=+
r2
qE
point charge Kq
1) Force, F=qE
-ne Q=Total charge E =
r Calculation of Charge tanθ= mg 1 r2
ε0=Permitivity of free space E
r2
Q1 r qE
n=1,2,3.... l
(
(
r1 r2
θ θ E
K=
1 x1 = 2) Acceleration, a= m
e=1.6 x 10-19C q q 4πε0 Q1 + Q 2
Q1 Q2 Sin θ= r
r
q
r
qE x1 x2 qE
2l 3) Velocity, V= t
2
< < N
Superposition Q1 Q2
m
Additivity of charge ε0 = [Q
[ ] 1
] [Q2]
[r2] [F]
=
[AT] [AT]
[L2] [MLT-2]
= M-1L-3T4A2 r = 2l sin θ mg r Q2 r
Q1 r1 2 E1
E x-
2= 4) Velocity, V= 2qE
Ql =Q1+Q2
( (
General rule
> net
x
>
Fair = q in equillibrium Q1 + Q2
r2 if θ is very small
F
Fmed= k Q2 ENet = m
Redistribution of charge Q1
F
k Q2 tanθ Sin θ θ
> E2 q2E2t2
Q2
r 5) Kinetic energy, K.E =
Q1
Unlike Charges 2m
r r k=dielectric constant of the medium r = qE E1
r1 2 2l mg ENet = E1 + E2
Q l=
Q1+Q2
Superposition q =-
( r1+r2 ( Q2 Q 1 in equillibrium
r /
k q2 r2
r3 ∝ q2
E2 Outside closer to smaller charge E E
-
+
2 Direction: = ,
E1
ENet = E1 - E2 accelerated in the accelerated opposite
Ql=Charge on each shell after redistribution a) Like- Towards the point at which force has to
2l mg Q1 Q2
E2
N direction of electric to the direction of
be evaluated (repulsion) field electric field
Charge Density b) Unlike- Away from the point at which force E1 r
r2 2 E
> net ENet =
(
Vy
( E12 + E22 +E1E2
>
Linear Charge density, λ=Q
C has to be evaluated (attraction) q =- Q1 Q 2 in equillibrium <|Q |
ELECTROSTATICS - 1
Unit= m r1+r2 If, E1=E2=E Then, Enet= 3 E 1 V = Vx 2+Vy 2
L F1
F
E V
General rule 60o
> E2
Q > net
>
q
Surface Charge density, σ =
( (
C
= U2+ qE t
2
Unit=
S m2 Vx = qEt
Density of ball “ρ” Q2 r m
(
u
(
x=
θ θ M
Q
Volume Charge density, ρ = Unit=
C θ
> F2 Enet
Q1 - Q2
>
V m3 accelerated in the direction of field and
>
F
F E1 ENet = E12 + E22 perpendicular to initial velocity
Q=Total charge V=Volume > net When θ =60o Distance from Q1=x+r
>
L=Length S=Area it θ does not change on submerging in liquid > If, E1=E2=E Then, Enet= 2 E
Fnet= 3F E2
MP
Dielectric constant of liquid, = 1837 , e = 1.7 10 -11
60o
+
>F m
Enet ENet = E12 + E22 -E1E2 Two point charges +8q and -2q are located Me 1 at2=h=Constant
E1 at x = 0 and x = L respectively.The location of a point
Fnet When θ =90o A charge is placed at the centre of 2
>
If, E1=E2=E Then, Enet= E
120O
>
on the x axis at which the net electric field due to E
p e
If a charge on the body is 1 nC,then 1 qE 2
the line joining two equal charges Q. The system ρ E2
(
(
θ θ
these two point charges is zero is:
how many electrons are present on the body? Fnet= 2 F t = h
2 m
F K= E
of the three charges will be in equilibrium if q ρ- Direction h
is equal to a) 8 L t1 h t2
a) 1.6 × 1019 b) 6.25 × 109 F
> t2 m
1) Positive charge:-Towards the point at which
tp m 1/2
[ [
b) 4 L
c) 6.25 × 1027 d) 6.25 × 1028 Fnet
density of liquid= electric field has to be evaluated = p
When θ =120o
a) -Q/2 c) +Q/4 c) 2 L te me
F Fnet= F 2) Negative charge:-Away from the point at which
b) -Q/4 d) +Q/2 electric field has to be evaluated d) L/4 tp>te
120O
F
3) Charge placed at the face
Time period of Charged Properties of field lines Electric flux Application of Gauss‛s Theorem 9) Electric field due to circular arc at its
center
Pendulum in an electric filed Flux is proportional to λ
Kq
1) Start from positive charge and end on total no. of field lines
passing through an area
q 1) Point charge E=
r2 E
++
++++ +++ ++
+++
Φ =
++ ++
++
negative charge s E
20
++
cube
QE l ∫
Φ = E.ds cos θ Qouter Qinner
T=2 π Φ =∫ E.ds Qinner= Qplate- Qouter
m)
(g- QE 2) Never intersect each other. If they intersect r
there will be 2 directions for electric field
2) Metal sphere/Hollow sphere
Eo= 2k
r
λ sin(θ/ )
2
mg which is not possible r
Gauss Law:- Φ = q0 = ∫0 E.ds cos θ
KQ
Tangent
E Time period will increase + +
P
4) Charge placed at the corner Esurface= +
electric lines
of force
Tangent
R2 +
R
+ E
+ Qplate θ
qnet KQ +
+
eg: For a semicircle θ=180o
E Zero flux:- Φ = = 0,where qnet=0 Φcube=
Q Eoutside= 2
r
r + +
+
R r
O
O
8 0
Eo= 2k λ sin(180/ )
0
l 3) P
o
180o
Always perpendicular to Q 1 Q r 2
T=2 π
++++
+++
Conducting surface Φone face= =
m)
(g+ QE Einside= 0
8 0 3 24
= 2k λ
++
QE Electric flux for Cube
+
++
0
++
r ++
++
++
mg 7) Electric field due to a finite linear ++
+++
++++++
++
1) No charge inside the cube 3) Non-Conducting sphere charge distribution
Time period will decrease 10) Electric field at the center of a
KQr circular ring
E α Electric field
+
4) Charge inside q=0 Einside= +
+
+++ ++++
E line density R3 + ++
++
q
+++ +++
+
l KQ E
+++ +++
QE 5) Charge placed at the edge O
T=2π E1
Φ= =0 Esurface= 2 λ +
Eo=0
++
R 2k λ +
++
r
(g2+( QEm )
2
E= r sin(θ/2) + (
mg
E1 > E2 0 θ
Time period will decrease KQ +
++
++
Eoutside= R r + +++ +++
E2
Charge inside q=0
Q r2 +
++
Φcube= +
4 0 11) Electric field due to a circular ring
Electric field inside a 2) Charge placed at the center Q Q +
Φface= 0 1 = 16 4) Conducting sheet of charge
dielectric medium 4
+
5)
0
Never form closed loops (Conservative force)
8) Electric field due to a infinite linear
q charge distribution E= π
4 ε
1 qx x2+r2
6) q α no. of field lines
Φ = 0
(x2+r 2)3/2
r
| q2| > | q1|
total
0 E= ∞
(For large distance)
0 2k λ θ
E= r x
Enet= E K q1 q2 6) Flux through curved surface
+
E= π
1 q
k +
+
4 ε0 x2
+
5) Non-conducting sheet
q
+
Electric lines of force about negative + r
λ
=
P
point charge are: Φ + E
a) circular, anticlockwise
one side
6 0 +
+ Emax
Φeffective = Φcurve+ 2Φcross section + + + + +
Enet=E-Einduced, Einduced=E-Enet =E (1- 1K )
b) circular, clockwise E= +
c) radial, inward 2 0
+ + + + +
x
x=r
d) radial, outward 2
