In equqation E=1/4PiEo 2pr/(r^2-l^2) .
If r is much greater than l .Value of l will be?
Answers
Answer:
Coulomb law
F = ke
Q1Q2
r
2
,ˆr (1)
ke =
1
4πo
= 8.98 × 109 Nm2
/C
2
(2)
o = 8.85 × 10−12C
2
/(Nm2
) (3)
E =
F
qo
(4)
For a point charge
E = ke
Q
r
2
ˆr (5)
For field lines
E ∝
# of lines
Area (6)
a =
qE
m
(7)
Gauss Law
ΦE =
Z
E · dA (8)
= E · A for constant field (9)
= EA cos(θ) for constant field (10)
ΦE = 4πkeQnet (11)
=
Qnet
o
(12)
1. Uniformly Charged slab.
E =
σ
2o
(13)
2. Two charged slabs
E =
σ
o
Inbetween the plates
0 Outside the plates (14)
3. The electric field from a long line of charge
Er =
2keλ
r
(15)
4. The electric field from a uniformly charged insulat-
ing sphere with total charge Q and radius R,
Er =
keQ
r
2 for r > R
keQ
R2
r
R
for r < R (16)
5. The electric field from a uniformly charged con-
ducting sphere with total charge Q and radius R,
Er =
keQ
r
2 for r > R
0 for r < R (17)
∆V = VB − VA = −
Z B
A
E · dx (18)
For a constant electric field Ez in the z direction
∆V = −E · d (19)
= −Ez d cos(θ) (20)
= −Ez z (21)
W
you
AB = UB − UA = q(VB − VA) (22)
or
∆U = q ∆V (23)
For a point charge
V (r) = keQ
r
(24)
Wyou = U = ke
q1q2
r12
+ ke
q2q3
r23
+ ke
q1q3
r13
(25)
Ez = −
∂V (z)
∂z (26)
Er = −
∂V (r)
∂r (27)
Ex = −
∂V
∂x Ey = −
∂V
∂y Ez = −
∂V
∂z (28)
V =
Xke
∆q
r
=
Z
ke
dq
r
(29)
1. Uniformly charged insulating sphere with charge Q
and radius R
V (r) = keQ
r
for r > R
keQ
2R