Charge Coulomb‛s Law Equillibrium of Charges Charge on pendulum Electric Field
PHYSICS
WALLAH Quantization of charge
Q1 F F Q2 F=
1
4πε0
Q 1Q 2
r2
E Electric field at a point
point charge Kq
Q=+ qE
-ne Q=Total charge E =
r Calculation of Charge tanθ= mg 1 r2
ε0=Permitivity of free space E
r2
n=1,2,3.... l
(
(
θ θ E 1
r1 r2 q
K=
4πε0
e=1.6 x 10-19C q
Q1 Q2 Sin θ= r r r
q 2l <
2
<
qE
Superposition
Additivity of charge [Q1] [Q2] [AT] [AT]
[ε0]= [r2] [F]
=
[L2] [MLT-2]
= M-1L-3T4A2 r = 2l sin θ mg
Q1 r1 2 E1
E
Ql =Q1+Q2
( (
General rule
> net
>
Fair = q in equillibrium
Q2 r2 if θ is very small
Fmed= k
ENet =
F F
Redistribution of charge Q1 k
r
Q2 tanθ Sin θ θ
> E2
Q1 Q2
r r k=dielectric constant of the medium r = qE E1
r1 2 2l mg ENet = E1 + E2
Q l=
Q1+Q2
2
Superposition
Direction:
q =-
( r1+r2 ( Q2 Q 1 in equillibrium
r = /
k q2 r2
r3 ∝ q2
E2
E1
Ql=Charge on each shell after redistribution 2l , ENet = E1 - E2
a) Like- Towards the point at which force has to mg E2
be evaluated (repulsion)
Charge Density b) Unlike- Away from the point at which force E1
r2 2 E
> net ENet =
( ( E12 + E22 +E1E2
>
has to be evaluated (attraction) q =- Q1
Linear Charge density, λ=Q Q 2 in equillibrium
C
ROSTATICS - 1
Unit= m r1+r2 If, E1=E2=E Then, Enet=
L F1
F General rule 60o
> E2
Q > net
>
Surface Charge density, σ = Unit=
C
S m2
Density of ball “ρ”
(
(
θ θ
Q
Volume Charge density, ρ = Unit=
C θ
> F2 Enet
>
V m3
>
F ENet =
Fnet E1 E12 + E22
Q=Total charge V=Volume > When θ =60o
>
L=Length S=Area it θ does not change on submerging in liquid > If, E1=E2=E Then, Enet= 2
Fnet= 3F E2
60o Dielectric constant of liquid,
>F Enet ENet = E12 + E22 -E1E
E1
Fnet When θ =90o A charge is placed at the centre of
>
120
>
O
If, E1=E2=E Then, En
If a charge on the body is 1 nC,then
the line joining two equal charges Q. The system ρ E2
(
(
θ θ
how many electrons are present on the body? F Fnet= 2 F
of the three charges will be in equilibrium if q K=
ρ- Direction
a) 1.6 × 1019 b) 6.25 × 109 > is equal to
F
1) Positive charge:-Towards the point at wh
density of liquid= electric field has to be evaluated
c) 6.25 × 1027 d) 6.25 × 1028 Fnet When θ =120o
a) -Q/2 c) +Q/4
F Fnet= F 2) Negative charge:-Away from the point a
120O
b) -Q/4 d) +Q/2 electric field has to be evaluated
F
3) Charge placed at the face
Time period of Charged Properties of field lines Electric flux Application of Gauss‛s Theor
Pendulum in an electric filed Flux is proportional to
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
negative charge s E Φ =
l ∫
Φ = E.ds cos θ 20
cube
QE T=2 π
m)
(g- QE 2) Never intersect each other. If they intersect ∫
Φ = E.ds
2) Metal sphere/Hollow sphere r
there will be 2 directions for electric field
mg which is not possible
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
+
qnet KQ +
+
E Zero flux:- Φ = = 0,where qnet=0 Φcube=
Q Eoutside= 2
r
r + +
+
R
0
8 0
l 3) Always perpendicular to Q 1 Q
P
T=2 π Conducting surface Φone face= =
m)
(g+ QE 8 0 3 24 Einside= 0
QE Electric flux for Cube
+
0
mg
3) Non-Conducting sphere
Time period will decrease 1) No charge inside the cube
KQr
E α Electric field Charge inside q=0
PHYSICS
WALLAH Quantization of charge
Q1 F F Q2 F=
1
4πε0
Q 1Q 2
r2
E Electric field at a point
point charge Kq
Q=+ qE
-ne Q=Total charge E =
r Calculation of Charge tanθ= mg 1 r2
ε0=Permitivity of free space E
r2
n=1,2,3.... l
(
(
θ θ E 1
r1 r2 q
K=
4πε0
e=1.6 x 10-19C q
Q1 Q2 Sin θ= r r r
q 2l <
2
<
qE
Superposition
Additivity of charge [Q1] [Q2] [AT] [AT]
[ε0]= [r2] [F]
=
[L2] [MLT-2]
= M-1L-3T4A2 r = 2l sin θ mg
Q1 r1 2 E1
E
Ql =Q1+Q2
( (
General rule
> net
>
Fair = q in equillibrium
Q2 r2 if θ is very small
Fmed= k
ENet =
F F
Redistribution of charge Q1 k
r
Q2 tanθ Sin θ θ
> E2
Q1 Q2
r r k=dielectric constant of the medium r = qE E1
r1 2 2l mg ENet = E1 + E2
Q l=
Q1+Q2
2
Superposition
Direction:
q =-
( r1+r2 ( Q2 Q 1 in equillibrium
r = /
k q2 r2
r3 ∝ q2
E2
E1
Ql=Charge on each shell after redistribution 2l , ENet = E1 - E2
a) Like- Towards the point at which force has to mg E2
be evaluated (repulsion)
Charge Density b) Unlike- Away from the point at which force E1
r2 2 E
> net ENet =
( ( E12 + E22 +E1E2
>
has to be evaluated (attraction) q =- Q1
Linear Charge density, λ=Q Q 2 in equillibrium
C
ROSTATICS - 1
Unit= m r1+r2 If, E1=E2=E Then, Enet=
L F1
F General rule 60o
> E2
Q > net
>
Surface Charge density, σ = Unit=
C
S m2
Density of ball “ρ”
(
(
θ θ
Q
Volume Charge density, ρ = Unit=
C θ
> F2 Enet
>
V m3
>
F ENet =
Fnet E1 E12 + E22
Q=Total charge V=Volume > When θ =60o
>
L=Length S=Area it θ does not change on submerging in liquid > If, E1=E2=E Then, Enet= 2
Fnet= 3F E2
60o Dielectric constant of liquid,
>F Enet ENet = E12 + E22 -E1E
E1
Fnet When θ =90o A charge is placed at the centre of
>
120
>
O
If, E1=E2=E Then, En
If a charge on the body is 1 nC,then
the line joining two equal charges Q. The system ρ E2
(
(
θ θ
how many electrons are present on the body? F Fnet= 2 F
of the three charges will be in equilibrium if q K=
ρ- Direction
a) 1.6 × 1019 b) 6.25 × 109 > is equal to
F
1) Positive charge:-Towards the point at wh
density of liquid= electric field has to be evaluated
c) 6.25 × 1027 d) 6.25 × 1028 Fnet When θ =120o
a) -Q/2 c) +Q/4
F Fnet= F 2) Negative charge:-Away from the point a
120O
b) -Q/4 d) +Q/2 electric field has to be evaluated
F
3) Charge placed at the face
Time period of Charged Properties of field lines Electric flux Application of Gauss‛s Theor
Pendulum in an electric filed Flux is proportional to
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
negative charge s E Φ =
l ∫
Φ = E.ds cos θ 20
cube
QE T=2 π
m)
(g- QE 2) Never intersect each other. If they intersect ∫
Φ = E.ds
2) Metal sphere/Hollow sphere r
there will be 2 directions for electric field
mg which is not possible
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
+
qnet KQ +
+
E Zero flux:- Φ = = 0,where qnet=0 Φcube=
Q Eoutside= 2
r
r + +
+
R
0
8 0
l 3) Always perpendicular to Q 1 Q
P
T=2 π Conducting surface Φone face= =
m)
(g+ QE 8 0 3 24 Einside= 0
QE Electric flux for Cube
+
0
mg
3) Non-Conducting sphere
Time period will decrease 1) No charge inside the cube
KQr
E α Electric field Charge inside q=0
