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# Solution of Electrostatics Test Paper 05 Class 12 Notes | EduRev

## Class 12 : Solution of Electrostatics Test Paper 05 Class 12 Notes | EduRev

``` Page 1

CBSE TEST PAPER-05
CLASS - XII PHYSICS (Unit – Electrostatics)
Ans 01.  Since
1
C
A o
C
d d
?
= ? ?
which means if distance increases, capacitance deceases.
Since
Q
V
C
= and charge on the capacitor is constant.
Hence reading of the voltmeter increases.
Ans 02. The introduction of dielectric in a capacitor reduces the effective charge on plate and
hence increases the capacitance.
Ans 03. The path of the electron traveling with velocity
m
s
? at right angles of  E

is
circular.
? It requires a centripetal force
2
mv
f
r
=
which is provided by an electrostatic force  f = eE
2
2
eE  =
=
mv
r
e v
m Er
?
Ans 04. For air
A o
C
d
?
=
Thickness   t  = d/2 only when k = 8
= =
1 1
(1 ) (1 )
2
o o
nef
A A
C
d
d t d
k
? ?
?
- - - - 8
o
2 A
=2C
d
?
=
Hence capacitance will get doubled.
Page 2

CBSE TEST PAPER-05
CLASS - XII PHYSICS (Unit – Electrostatics)
Ans 01.  Since
1
C
A o
C
d d
?
= ? ?
which means if distance increases, capacitance deceases.
Since
Q
V
C
= and charge on the capacitor is constant.
Hence reading of the voltmeter increases.
Ans 02. The introduction of dielectric in a capacitor reduces the effective charge on plate and
hence increases the capacitance.
Ans 03. The path of the electron traveling with velocity
m
s
? at right angles of  E

is
circular.
? It requires a centripetal force
2
mv
f
r
=
which is provided by an electrostatic force  f = eE
2
2
eE  =
=
mv
r
e v
m Er
?
Ans 04. For air
A o
C
d
?
=
Thickness   t  = d/2 only when k = 8
= =
1 1
(1 ) (1 )
2
o o
nef
A A
C
d
d t d
k
? ?
?
- - - - 8
o
2 A
=2C
d
?
=
Hence capacitance will get doubled.
Ans 05.
2
1
2
U CV =
For parallel plate
2
1
2
A o
U V
d
?
=
When d’ = d – 10% of d  == 0.9 d
Then
2
1
'
2 0.9
A o
U V
d
?
=
Change in energy = U’-U =
2
1 1
( 1)
2 0.9
AEo
V
d
- 0.1
' ` ( )
9
0.9
U
U U U - = =
% change ==
' 1 100%
100 %=    X 100%=
9
9
U U
U
U U
- × ×
% change == 11.1%
Ans 06. (a) Since
2
10
1000
.
1 10
d V
V
E
m
dr
x
?
- = = =
(ii)  Since surface A is an equipotential  surface ie 0 V ? =
? Work done from X to Y  = Zero .
(b)
if  E=0
0  dv = 0 or V= constant (non zero).
dV
E
dr
dV
dr
- =
= ?
Ans 07. (a) We know
1
V
r
a
Vp >  Vp -- Vq  -- Positive Vq?
B A
VA < V -- V  -- Positive
B
V ?
Because V B    is less negative then V A.
Page 3

CBSE TEST PAPER-05
CLASS - XII PHYSICS (Unit – Electrostatics)
Ans 01.  Since
1
C
A o
C
d d
?
= ? ?
which means if distance increases, capacitance deceases.
Since
Q
V
C
= and charge on the capacitor is constant.
Hence reading of the voltmeter increases.
Ans 02. The introduction of dielectric in a capacitor reduces the effective charge on plate and
hence increases the capacitance.
Ans 03. The path of the electron traveling with velocity
m
s
? at right angles of  E

is
circular.
? It requires a centripetal force
2
mv
f
r
=
which is provided by an electrostatic force  f = eE
2
2
eE  =
=
mv
r
e v
m Er
?
Ans 04. For air
A o
C
d
?
=
Thickness   t  = d/2 only when k = 8
= =
1 1
(1 ) (1 )
2
o o
nef
A A
C
d
d t d
k
? ?
?
- - - - 8
o
2 A
=2C
d
?
=
Hence capacitance will get doubled.
Ans 05.
2
1
2
U CV =
For parallel plate
2
1
2
A o
U V
d
?
=
When d’ = d – 10% of d  == 0.9 d
Then
2
1
'
2 0.9
A o
U V
d
?
=
Change in energy = U’-U =
2
1 1
( 1)
2 0.9
AEo
V
d
- 0.1
' ` ( )
9
0.9
U
U U U - = =
% change ==
' 1 100%
100 %=    X 100%=
9
9
U U
U
U U
- × ×
% change == 11.1%
Ans 06. (a) Since
2
10
1000
.
1 10
d V
V
E
m
dr
x
?
- = = =
(ii)  Since surface A is an equipotential  surface ie 0 V ? =
? Work done from X to Y  = Zero .
(b)
if  E=0
0  dv = 0 or V= constant (non zero).
dV
E
dr
dV
dr
- =
= ?
Ans 07. (a) We know
1
V
r
a
Vp >  Vp -- Vq  -- Positive Vq?
B A
VA < V -- V  -- Positive
B
V ?
Because V B    is less negative then V A.
(b) In moving a positive change form Q to P work has to be done against the
electric field so it is negative.
(c) In moving a negative change form B to A work is done along the same direction
of the field so it is positive.
Ans 08.  Vandegraff generator is a device which is capable of producing a high potential of
the order of million volts.
Principle (1)  The charge always resides on the outer surface of hollow conductor.
(2) The electric discharge in air or gas takes place readilly at the pointed
ends of the conductors.
Construction:- It consist of  a large hollow
metallic sphere S mounted on two
insulating columns A and B and an endless
belt made up of rubber which is running
over two pulleys P 1 and P 2   with the help of
an electric motor.  B 1 and B 2 are two sharp
metallic brushes.  The lower brush B 1 is
given a positive potential by high tension
battery and is called a spray brush while
the upper brush B 2 connected to the inner
part of the Sphere S.
Working :- When brush B 1 is given a high positive potential then it produces ions, due to the
action of sharp points.  Thus the positive ions so produced get sprayed on the belt due to
repulsion between positive ions and the positive charge on brush B 1.  Then it is carried
upward by the moving belt.  The pointed end of B 2 just touches the belt collects the positive
change and make it move to the outer surface of the sphere S.  This process continues and the
potential of the shell rises to several minion volts.
Applications – Particles like proton, Deutrons, a -- particles etc are accelerated to high
speeds and energies.
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