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Electrostatics: Electric
Charges and Fields Class 12
Physics Chapter 1 Notes
JEE Mains & Advanced
Get complete Electrostatics: Electric Charges
and Fields class 12 physics chapter 1 notes.
Electric Charge is that physical property of
matter due to which the other matter experience
a force when matters are placed in
electromagnetic field. Electric charge is also
known as Charge, Electrical Charge and
Electrostatic Charge. It is denoted by symbol
‘q’. It is a scalar quantity as charge has only
magnitude and no direction.The two types of
charges exist in nature: Positive and Negative
Charge. Here are the detailed Physics Electric
Charges and Fields Class 12 Notes . These
notes provide an exclusive insight of
electrostatics. Thus the notes will be very
helpful in preparation for entrance examinations
like IIT JEE & NEET.
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Electrostatics: Electric Charges and Fields
class 12 physics chapter 1 notes
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Electrostatics: Electric Charges and
Fields class 12 physics chapter 1
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ELECTROSTATICS
(v)
Induction
takes
place
only
in
bodies
(either
conducting
or
non
conducting)
and
not
in
particles.
(c)
Conduction:
When
an
insulated
conductor
is
brought
in
contact
with
a
charged
body
and
it
gets
the
same
charge
as
the
charged
body
then
conduction
takes
place.
Conduction
is
only
possible
in
conductors
and
not
in
insulators.
EXAMPLES
BASED
ON
CHARGE
Ex.1
Two
metallic
spheres
same
in
all
respects
are
taken.
One
is
given
a
positive
charge
of
Q
coulomb
and
other
is
given
the
same
but
negative
charge.
Which
sphere
will
have
a
higher
mass.
Sol.
Negatively
charged
sphere
will
have
a
higher
mass.
This
is
due
to
increase
in
number
of
electrons
to
make
it
negatively
charged.
Ex.2
Which
of
the
followi ng
charge
is
not
possible-
[1]
1.6
×
10-18
C
[2]
1.6
×
10-19
C
[3]
1.6
×
10-20
C
[4]
None
of
these
Sol.
1.6
*
10-20
C,
because
this
is
1/10
of
electronic
charge
and
hence
not
an
integral
multiple.
Ex.3
How
many
electrons
are
present
in
1
coulomb
charge.
Sol.
:
q
=
ne
q=
1C
e
=
1.6
×
10-19
C
So
n
=
q/e
=
6.25
x
1018
electrons.
Ex.4
Identify
X
in
the
followi ng
nuclear
reactions-
[a]
'H
+
9Be
→
X
+
n
[b]
12C
+
1H
→
X
[c]
15N
+
1H
→
*He
+
X
Sol.
Using
conservation
of
charge
[a]
Charge
on
nucleus
of
X
must
be
(because
Q
(n)
=
0).
=
Q
(H)
+
Q
(Be)
=
1e
+
4e
=
5e
Thus
X
is
boron
(B).
[b]
Q(X)
=
Q(C)
+
Q(H)
=
6e
+
1e
=
7e
Thus
X
is
nitrogen
(N).
[c]
Q(X)
+
Q(He)
=
Q(N)
+
Q(H)
Q(X)
+
2e
=
7e
+
1e
or
Q(X)
=
6e
Thus
X
is
carbon
(C)
4.
COULOMB'S
LAW
The
force
of
attraction
or
repulsion
between
two
stationary
point
charges
is
directly
proportional
to
the
product
of
charges
and
inversely
proportional
to
the
square
of
distance
between
them.
This
force
acts
along
the
line
joining
the
centre
of
two
charges.
If
a,
&
92
are
charges,
r
is
the
distance
between
them
and
F
is
the
force
acting
between
them
Then,
Fa
q,
92,
F
a
1/12
..
Fa
9192
5
ELECTROSTATICS
+q
+q
•
B
*Q+q
さ
+
•
B
Q
=
0
Q
=
positive
Q=
-q
(A)
(B)
(C)
Now
if
the
outer
sphere
is
given
a
charge
Q
or
is
earthed,
the
result
of
case
(a)
and
(b)
will
not
change
as
external
charge
on
a
conductor
does
not
affect
the
field
inside
it.
However,
in
case
(c),
i.e.,
r
>
b.
i)
If
the
outer
sphere
is
given
a
charge
Q,
the
charge
Q
will
also
produce
electric
field
so
that
now
:
Ec
=
Ec+
4780
［
>
E
d
This
is
shown
in
fig.
(B).
(ii)
If
the
outer
sphere
is
earthed,
the
outer
sphere
will
acquire
induced
charge
say
q'
such
that
its
potential
becomes
zero
as
potential
of
earthed
conductor
is
assumed
to
be
zero
then
V
=
-
1(9+a)
=0
i.e.,
q'=-a
4n80
i.e.,
the
induced
charge
on
the
outer
shell
will
now
be
(-q)
[and
not
zero]
so
now
Ec
=Ec+
1(9-9)
=0
4
T
E
。
K
Ed
i.e.,
field
outside
the
shell
is
zero.
In
this
situation
field
exists
only
in
the
space
between
the
shells
as
shown
in
fig.
(C).
Ex.2
A
conducting
hollow
spherical
shell
having
an
inner
radius
a
and
outer
radius
b
carries
a
net
charge
Q.
If
a
point
charge
q
is
placed
at
the
centre
of
this
sphere,
determine
the
surface
charge
density
at
(a)
the
inner
surface
and
(b)
the
outer
surface.
Sol.
If
ga
and
a
are
the
charges
on
the
inner
and
outer
surfaces
of
the
given
shell
96
G
a
=
Ana?
and
op=
4
T
b
？
The
isolated
sphere
is
given
a
charge
Q,
so
in
accordance
with
conservation
of
charge
9
+
9,
=
Q
Intensity
at
a
point
inside
a
conductor
is
zero
and
external
charge
does
not
produce
field
inside
a
conductor,
so
for
point
P
(a
<
r
<
b),
we
have
1
(q+
9a
Ep
=
=
0,
i.e.,
9a
=
-9
4
T
E
o
P
So
that
Eqn.
(2)
yields
96
=
Q
-
9a
=
Q
-
(-9)
=
Q+
a
And
hence
in
the
light
of
eqn.
(1)
(a)
Ca
=
4па?
and
(b)
Q+9
4
T
b
？
12
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