10 Electrostatics
1.
In the following four situations charged particles are are at equal equal distan distance ce from from the origin origin.. Arrang Arrange e them the magnitude of the net electric eld at origin greatestY rst 5q
(i)
(ii) 4%q
#q
Y
(iii) O
'q
(d) Independe Independent nt of charge charge
X
O
#q
(c) ore in the the case case of proton proton 5.
Y
4 %q
(b) ore in in the case case of an electron electron
X
O
4q
X
4 #q
(a) ero
Y
(iv) %q
O
4q
(c) (c) (i) (i) > (ii (iii) i) > (ii) (ii) > (iv) (iv) (d) (d) (iv) (iv) > (ii (iii) i) > (ii) (ii) > (i) (i) A linear linear char charge ge having having linear linear char charge ge densit density y λ , penetrates a cube diagonally and then it penetrate a sphere diametrically as shown. hat will be the ratio of !u" coming cut of cube and sphere
q
:
:
(d) $#q# 'πε r 6.
:
:
:
q
q
0ive point charge each having magnitude 1 q2 are placed at the corner of he"agon as shown in g. electric eld at 1O2 be
/E
E
. &o get net
, charge placed on the
q remaining si"th corner should be
q
(a) / q
a :
q
3et electric eld at the centre 1O2 is
: : : : : : : :
q
r
(c) q# 'πε r
: :
q
(b) /q# 'πε r
X
(a) (i) > (ii) (ii) > (iii) (iii) > (iv) (iv) (b) (ii) (ii) > (i) (i) > (iii) (iii) > (iv) (iv) 2.
A point charge is surrounded symmetrically by si" ident identica icall charge charges s at distan distance ce r as shown in the gure. gure. -ow much much wor* is done by the the forces forces of electrostatic repulsion when the point charge q at the centre is removed at innity
(b) 4 / q
:
a
q
O
(c) 5 q
q
q
(d) 4 5 q (a) (c) 3.
$
(b)
# %
(d)
#
#
7.
%
$ $
&wo &wo charges each equal to η q(η $ < %) are placed placed at the corners corners of an equilater equilateral al triangle of −
side a. &he electric eld at the third corner is E% where (E
4.
= q 'πε a#)
(a) E %
= E
(b) E%
< E
(c) E%
> E
(d) E %
≥ E
An elec electr tron on falls falls thro throug ugh h a small small dist distan ance ce in a −$ . ' uniform uniform electric eld of magnitude magnitude
# × $ NC
&he direction direction of the eld is reversed reversed *eeping *eeping the magnitude magnitude unchanged unchanged and a proton proton falls through through the same distance. &he time of fall will be (a) +ame +ame in both both cases cases
8.
An innit innite e non6co non6cond nduct ucting ing sheet sheet has a surfa surface ce # charge charge density density σ 7 7 .$ µ Cm on one side. -ow far far apar apartt are are equi equipo pote tent ntia iall surf surfac aces es whos whose e potentials di8er by 5 V (a) (a) 9.95 .95 m
(b) (b) 9.95 9.95 cm
(c) (c) 9.95 .95 mm
(d) (d) 99.5 99.5 mm
0igure shows a charged conductor resting on an insu insula lati ting ng stan stand. d. If at the the poin pointt P the charge charge densit density y is
σ , the poten potentia tiall is V and and the electric
eld eld stren strength gth is E, what are the values of these quantities at point Q
P
Q
Insulating stand
11 Electrostatics
;harge density
=lectric intensity
> σ > σ < σ < σ
> V
>E
V
>E
V
E
V
E
(a) (b) (c) (d) 9.
(c) x =
11.
a # #a # −$
(b) x =
(d) x =
(b) Beft ward at constant speed (c) Accelerated right ward (d) Accelerated left ward 14.
(b) %
(c) '
(d) /
→
p
P
4Q
(a)
k . p x %
15.
→
p
→
p
:Q
(b)
(d)
#kp
x % # kp
x %
In the following diagram, the charge and potential di8erence across 9 µ F capacitance will be respectively C µ F
→
B
@ µ F %/ µ F / µ F ' µ F 9 µ F @ µ F
E A
C
' V D
In moving from A to B along an electric eld line, the electric eld does /.' × $−$@ J of wor* on an electron. If φ $, φ # are equipotential surfaces, then the potential di8erence (V C
− V A ) is
B
16.
(a) %# µ C, ' V
(b) '# µ C, 5 V
(c) #$' µ C, #C V
(d) %/ µ C, '5 V
A conducting sphere of radius R, and carrying a charge q is Doined to a conducting sphere of radius #R, and carrying a charge 4 # q. &he charge !owing between them will be (a)
E
(b) 'V
q %
(b)
C
A
(c) ero (d) /' V
:Q
x
x
(c) ero
(d) D
(a) 4 'V
:Q
4Q
# +$
(c) C
12.
4 ' V
x
&he electric eld in a region surrounding the origin is uniform and along the x-a"is. A small circle is drawn with the centre at the origin cutting the a"es at points A, B, C, D having co6ordinates (a, ), (, a), (4 a, ), (, 4 a)? respectively as shown in gure then potential in minimum at the point
(b) B
4 Q 'πε $
#a
(a) A
4 # V
&hree identical dipoles are arranged as shown below. hat will be the net electric eld at P
k =
#a
&wo identical balls having li*e charges and placed at a certain distance apart repel each other with a certain force. &hey are brought in contact and then moved apart to a distance equal to half their initial separation. &he force of repulsion between them increases '.5 times in comparison with the initial value. &he ratio of the initial charges of the balls is (a) #
In the following gure two parallel metallic plates are maintained at di8erent potential. If an electron is released midway between the plates, it will move (a) ight ward at constant speed
&wo point charge 4 q and :q/ # are situated at the origin and at the point ( a, , ) respectively. &he point along the X-a"is where the electric eld vanishes is (a) x =
10.
13.
(c) q φ $
φ #
(d)
#q % 'q %
Electrostatics 17.
An arc of radius r carries charge. &he linear density of charge is λ and the arc subtends a angle π
%
22.
at the centre. hat is electric potential at the
−
λ
(b)
'ε λ
(c) 18.
19.
@
co!om , then the potential of inner
sphere will be, if the outer sphere is earthed
λ
(a)
&he radii of the inner and outer spheres of a condenser are @cm and $cm respectively. If the dielectric constant of the medium between the two spheres is / and charge on the inner sphere is $9× $
centre 9ε λ
(d)
$#ε
23.
$/ε
(a) $9 #o!$
(b) % #o!$
(c) $9 #o!$
(d) @ #o!$
In the following gure, the charge on each condenser in the steady state will be
(a) ' m
(b) ' mm
(c) ' µ m
(d) ' pm
%Ω
'Ω
$Ω $V
:Q
(b)
(a) % µ C
(b) / µ C
(c) @ µ C
(d) $# µ C
In the circuit, shown in g. 1 % 2 is open. &he charge on capacitor C in steady state is q$. 3ow *ey is closed and at steady state, the charge on C is q#. &he ratio of charges R
21.
:Q
E
(d)
(a)
A 5 µ F capacitor is charged at a steady rate of $ µ C/sec. &he potential di8erence across the capacitor will be $ V after an interval of (a) 5 sec
(b) # sec
(c) #5 sec
(d) 5 sec
A
: B
4
(a) /C (b) 5C
C
C
C
(c) %C (d) #C
C
C
% #
(c) $ 25.
0ind equivalent capacitance between A and B C
% µ F
/Ω /Ω
:Q
(c)
/Ω
% µ F
24.
:Q
% µ F
%Ω
A charge Q is "ed at a distance d in front of an innite metal plate. &he lines of force are represented by
(a)
%Ω
% µ F
A neutral water molecule (H#O) in itEs vapor state has an electric dipole moment of magnitude /.' 4% ×$ C-m. -ow far apart are the molecules centres of positive and negative charge
:Q
20.
12
q$ is q# % C
#R
(b)
(d)
# %
$ #
In a parallel plate capacitor the separation between the plates is %m with air between them. 3ow a $mm thic* layer of a material of dielectric constant # is introduced between the plates due to which the capacity increases. In order to bring its capacity to the original value the separation between the plates must be made (a)
$.5 mm
(b) #.5 m
(c)
%.5 mm
(d)
'.5 mm
13 Electrostatics
(SET -18)
Y
=
+ince
X
X
O
(iii) #
+
#
(%E)
#
(5E)
=
#
(%E)
+
= ' E + #E = /E E(&# ) = %E + E = 'E ⇒ E(&) > E(&&&) > E(&&) >
=
4.
(c)
%E E
cube φ $
=
ε
and
5 #E , 5.
% #E ,
∴
φ #
=
λ ⋅ #a ε
......(ii)
φ $ φ #
%
=
#
(c)
3.
E$ = E
η q #
'πε a
= E$ +
E#
, E#
=
η q #
'πε a
.
distance
.
%η :> q $.
%η q 'πε a
#
>
:q
q # 'πε a
⇒
&herefore
6.
d
is
d =
$ qE
$ # # m
or
#dm
qE
+ince $ # (b)
∝ m, a proton ta*es more time. &otal
potential
at
the
/q 'πε r
equired wor* done
E(&# )
sphere
E
#
'πε a
&he time required to fall
centre V =
and
......(i)
from
(c)
$ =
0lu" coming out of the λ . a %
O %η , E
through
X
E(&&&)
2.
%, $ <
:q
(iv)
(5E)
E
%η q
q . E % > E E = # 'πε a
Y
'E #E
=
4 5
η
X
#E E
(ii)
Y
E(&&)
$
−
⇒
O
(i)
=
+ E + #E$ E # cos/o = 5E E
5E
E(&)
E
# #
Y
#E %E
O
# $
E
%E O
'q
'q
(c) If electric eld due to charge FqF at origin is E then electric eld due to charges F#qF, F%qF, F'qF and F5qF are respectively #E, %E, 'E and 5E
1.
= q.V =
/q
#
'πε r
(d) &o obtained net eld /E at centre O, the charge to be placed at remaining si"th corner is 4 5 q. (see following gure)
14
Electrostatics
7.
(c) E =
=
σ V ⇒ #ε d
V V × #ε ⇒ d = d σ
=
−$#
5× #× 9.95× $ −/
.$× $
11.
(d) &he surface of the conductor is an equipotential surface since there is free !ow of electrons within the conductor. &hus potential at Q is the same as that at P. &hat is V P = V Q = V . &he electric eld E at a point on the equipotential surface of the conductor is inversely proportional to the square of the radius of curvature r at that point. &hat is −#
E ∝ r
12.
(c)
(,, )
*
13.
kq #
x
=
kq # ( x − a)
(V C
E$
= x #
=
10.
( #
−
$) x =
E#
or x =
#a
# − $ #a
15.
Q#
A
B
k . p
Q$ + Q# #
Q$ + Q# #
A
=
and
k . p
is
given
k (Q$ + Q#)# r #
r #
F E = '.5F
= '.5 k .
Q$Q# r #
:Q
E#
E$
E%
=
:Q
4Q
P
%
E%
#
:Q
k .(# p) (towards x %
V $ V #
= 'V
/ µ F
%/
' µ F
=
$9
=#
%/ µ F
9 µ F
and V # =
k (Q$ + Q#)#
that
$
4Q
x #
#
It
4Q
(towards left)
-ence V $ =
r
Q$ + Q# # = F E = # r #
e
right) +o net eld at P will be ero. (c) Jiven circuit can be redrawn as follows capacitors, @ µ F , @ µ F and C µ F are short circuited. +o they are deleted.
B
k
/.' × $−$@
(towards left)
x %
V $ + V #
r
0inally G
= ( =
V A )
due to dipole %
(a) +uppose the balls having charges Q$ and Q# respectively. Initially G Q$
−
due to dipole #
or
#( x − a) = x
or
( V B
due to dipole $
or #( x − a)#
#
−e(V A − V B ) − V A ) 7 e(V C − V A )
-ence eld at P
x
+uppose the eld vanishes at a (distance x ), we have
.
$
= 'V $./ × $−$@ (d) =lectric eld is directed right ward (higher potential of 4 # V to lower potential of 4 ' V ) . hen electron left free in an electric it accelerates opposite to the electric eld. -ence in the given case electron accelerates left ward. (c)
N
(a,, x )
it
7 V C)
14.
q/ #
#
Hn solving
(a) In the direction of electric eld, potential decreases. (b) or* done by the eld 7 e(V B
+
)q
=
= '.5 Q$Q# .
( = q(−dV ) =
+ince point Q has a larger radius of curvature than that at point P, the electric eld at Q is less than that at P. &hat is E Q < E P = E 9.
Q$ Q#
gives
7 9.95 × $4% m 7 9.99 mm 8.
(Q$ + Q#)#
⇒
9 %
' %
V
V $
V #
' V
V
;harge on 9 µ F capacitor 9 = 9× = #$%.% µ F ≈ #$'µ F %
so 16.
(d) Initial charge on sphere of radius R 7 q ;harge on this sphere after Doining
qE =
(q + (−#q)× R
R + #R
=
−q × R %R
=−
q %
15 Electrostatics
3ow charge !owing between them
q = q − − = % 17.
'q %
= r θ =
(c) Bength of the arc ;harge on the arc ∴
= 18.
$ 'πε
r π
=
=
=
r π !λ
%
× λ
%
at center
r π λ % r
×
3ow potential of inner sphere will be equal to potential di8erence of the capacitor. +o q $9× $−@ = V = −$ = %V C / × $ 23.
r
r
kq r
/ % π
V = #× '
λ
***
$#ε
(c) &here are $ electrons and $ protons in a neutral water molecule. +o itEs dipole moment is p 7 q (#!) 7 $ e (#!) -ence length of the dipole &,e, distance between centres of positive and negative charges is
#!
=
p $e
=
−#
/.' × $
$× $./ × $−$@
−$#
= ' × $
m7
20.
(a) etal plate acts as an equipotential surface, therefore the eld lines should enter normal to 25. the surface of the metal plate. (d) ;harge required to reach the capacitor upto $ V is
Q
= 5× $−/ × $ = 5 × $−% C
3ow required time
21.
=
5 × $−%
7 5 sec $× $− / (d) All capacitor lying in left side of line XY are short circuited so circuit can be reduced as follows C
X
C
A )
C
'
A
B
)
'
B
C
C ⇒
C
C
C Y
C AB 22.
= #C
(b) Jiven system is a spherical capacitor +o
capacitance
of
system
r # − r $
C = % × 'πε
=
/ @
@ × $
24.
r $r #
@ × $ × $− # = / × $−$ Farad $
=
9 #o!$
-ence, potential di8erence across each capacitor is 'V +o charge on each capacitor Q 7 % × ' 7 $# µ C. (a) hen *ey is open, charge in steady state will be q$ = CE. hen *ey is closed, potential di8erence across #R # E= R capacitor will be V = R + #R % ;harge in steady state will be q# =
q$
' pm 19.
(d) In steady state current !ows through 'Ω $ = #amp. resistance only and it is & = (' + $)
q#
=
(c) % =
# %
CE
% #
.
$ $ − d E
⇒
+o new distance
#=
$ $ − d E
= %+
$ #
⇒
d E =
= %.5 m
$ #
m
⇒
