Download Electric lines of force do not intersect!

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Ch.Rama Krishna
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AMøtºÆŠæ l 8 l 2015
Electric lines of force do not intersect!
ELECTROSTATICS
Electric Flux:
The space around the charge up
to which it influence can be felt
is called electric field. It can be
expressed in terms electric lines
of force called electric flux.
The scalar
product of electric
field (E ) and small
element of
surface area (∆S) is called
electric flux ∆φ passing
through
the area element (∆S)
Consider a point P which is at a
distance r from a point charge q.
Imagine a spherical Gaussian surface of radius r with its centre
coinciding with the charge q.
Then according to Gauss theorem :
1
∫ E ⋅ ds = ε0 q
q
∫ E ⋅ dS = ε0
Applications of Gauss law:
1) Electric field due to point
charge (q):
+q
r
P
Important questions for IPE
1. The electric lines of force do not
intersect. Why?
2. State Gauss's law in electrostatics.
3. When is the electric flux negative and when is it positive?
4. Define electric flux. Applying
Gauss's law and derive the expression for electric intensity due
to an infinite long straight charged wire. [Assume that the electric
field is everywhere radial and
depends only on the radial distance 'r' of the point from the wire.]
5. State Gauss's law in electrostatics. Applying Gauss's law derive
the expression for electric intensity due to an infinite plane sheet
of charge.
6. Applying Gauss's law derive the
expression for electric intensity
due to a charged conducting spherical shell at (i) a point outside
the shell (ii) a point on the
surface of the shell and
(iii) a point inside the shell.
JEE MODEL QUESTIONS
1. Two non-conducting solid spheres of radii R, 2R, having uniform volume charge densities ρ1
and ρ2 respectively, touch each
Physics
Physics
E2 πr =
ds
∫ E ⋅ dS
0
Sr. Inter
q
E4 πr =
ε0
2
2) Electric field due to infinite
long uniformly charged wire:
E
Gauss Law :
The total electric flux passing through any closed surface is equal to
1
times the total charge(q)
ε
enclosed by the closed surface.
1 q
E=
4πε 0 r 2
E
∆φ = E ⋅ ∆ S or φ =
ds
+
+
+
+
+
+
+
+
+
+r
+
+
+
+
+
+
E=
E=
l
λ
ε0
λ
2 πε 0 r
Imagine a symmetrical cylindrical Gaussian surface of l and area
of cross section ds
The electric field is perpendicu lar to flat surface. i.e. E , ds are
parallel, for a curved surface E
, ds are perpendicular to each
other.
According to Gauss law:
q
σds
∫ E ⋅ ds = ε0 2Eds = ε0
σ
σ
E=
E=
nˆ
2ε 0
2ε 0
where n is a unit vector normal
to plane and going away from it.
Electric field due to charged
conducting shell:
Outside the Shell (r > R):
λ
nˆ
2 πε 0 r
Let λ be linear density of the uniform charged wire i.e.,
Imagine a cylindrical Gaussian
surface of length l and radius r
with its axis coinciding with the
infinite charged long wire.
Since field is radial flux passing
through flat surface is zero, flux
passing through curved surface
only.
According to Gauss theorem:
q
∫ E ⋅ dS = ε0
E
32
c) 25
+σ
++++
++++
++++
++++
++++
ds
ds
E
ds
Let q be charge which distributed
uniformly over surface area of
the hallow sphere of radius R.
Surface charge density
σ=
E
Let σ is surface charge density
i.e.
σ=
other. The net electric field at a
distance 2R from the centre of
the smaller sphere, along the line
joining the centre of the spheres
ρ1
is zero. The ratio ρ can be
2
a) – 4
−32
b) 25
where n̂ is radial unit vector in
the plane normal to the wire.
Electric field due to an infinite
plane sheet of charge:
Q total charge
=
A
total area
q
4πR 2
Consider a point P outside the
shell at a distance r from the
centre of the shell.
Consider a Gaussian spherical
surface of radius r such that its
centre coincides with centre of
| E(r) |
at (0, 0, 0) and -q at
a 

 0, + ,0 
4 

V(r)
E4 πr 2 =
q
ε0
Electric field due to a charged
spherical shell:
At inside point r < R:
Consider a charged conducting
spherical shell of surface charge
density σ =
r
Q
λ=
L
q en
ε0
q
4πε 0 r 2
E=
∫ E ⋅ dS =
E
the shell.
According Gauss law:
q
4πR 2
P is a point inside the shell at a
distance 'r' from centre of shell.
Imagine a Gaussian surface of
radius 'r' such that its centre coincides with the centre of the shell.
According to Gaus theorem,
∫ E ⋅ dS =
q enclosed
0
E4 πr 2 =
ε0
ε0
Since charge enclosed by Gaussian surface is zero. E = 0
Thus the field due to a charged
conducting shell is zero at all
points inside the shell.
E
r
R
The value of k is …
a)
Choose the correct option(s)
0
b)
r
R
| E(r) |
V(r)
d) 4
2. Two non-conducting spheres of
radii R1, R2 and carrying uniform volume charge densities +ρ
–ρ, respectively, are placed such
that they partially overlap, as shown in the figure. At all points in
the overlapping region,
0
a) The net electric flux crossing
+a
the plane x = 2 is equal to
the net electric flux crossing
−a
the plane x = 2
b) The net electric flux crossing
+a
the plane y = 2 is more than
the net electric flux crossing
−a
the plane y = 2
c) The net electric flux crossing
q
the entire region is ε .
d) The net electric flux crossing
+a
the plane z = 2 is equal to
the net electric flux crossing
+a
the plane x = 2
4. Consider a thin spherical shell of
radius R with its centre at the
origin, carrying uniform positive
surface charge density. The variation of the magnitude of the
electric field E(r) , the electric
potential V(r) with the distance r
from the centre, is best represented by which graph ?
R
| E(r) |
r
V(r)
c)
6. Six point charges are kept at the
vertices of a regular hexagon of
side L and center O, as shown in
the figure. Given that ,
K=
0
R
| E(r) |
r
V(r)
1 q
4πε 0 L2
Which of the following statement(s) is/are correct?
d)
0
R
r
0
a) the electrostatic field is zero
b) the electrostatic potential is
constant
c) the electrostatic field is constant in magnitude
d) the electrostatic field has same direction
3. A cubical region of side a has its
centre at the origin. It encloses
three fixed point charges, -q at
a 

 0, − ,0  , + 3q
4 

5. An infinitely long solid cylinder
of radius R has a uniform volume charge density ρ. It has a
R
spherical cavity of radius
2
with its centre on the axis of
the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is
at a distance 2R from the axis of
the cylinder, is given by the
expression
23ρR
.
16kε 0
a) The electric field at O is 6 K
along OD
b) The potential at O is zero
c) The potential at all points on
the line PR is same
d) The potential at all points on
the line ST is same
Answers
1) b, d
4) d
2) c, d
5) 6
3) a, c
6) a, b, c