Download SOLUTIONS OF OBJECTIVE TEST GAUSS`S LAW AND ELECTRIC

SOLUTIONS OF OBJECTIVE TEST GAUSS’S LAW AND ELECTRIC POTENTIAL [SET A] FOR SET B SEE SOLUTIONS
1. (b) The work done by the conservative field is independent of the path followed.
Field,
= Displacement, = =-a - b ). W = (q
= - (q E ).
(a + b )
= -q a E.
2. (c) The electric flux through a Gaussian surface is only due to the charges present inside the Gaussian surface.
But the electric field on the Gaussian surface is due to the charges present both inside and outside the
Gaussian surface. It will be due to all the charges.
3. (b) The fields due to the three sheets at point P will point in the negative Z-direction.
Total electric field at point P is
(- )+
(- )+
(- )
=.
=
4. (d) Flux through each face of the cube,
ф E= X
!
=
"
.
5. (b) For points inside a charged sphere,
Eαr
And for points outside a charged sphere,
Eα
Hence option (b) is correct.
$
6. (a) Inside the inner spherical shell A,E=0. At r=rA, the field is constant. As QB is –ve and |QB|>| QA|, the field in
the space between A and B decreases with the increase in the value of x and become constant at the
surface of B. Outside the shell B, the field E decreases from a negative value to zero. Hence option (a) is
correct.
7. (d) When charge q is placed at one corner, the flux through each of the three faces meeting at this corner will
be zero , as is parallel to these faces. One-eighth of the flux emerging from charge q passes through the
remaining three faces, so the flux through each such face is
"
% & фE= . .
=
"
. .
' 8. (b) Electric potential at any point inside a hollow sphere is same as that on its surface. As the potential at the
surface is 10 V, potential at the center will also be 10 V.
9. (a) For a spherical conductor,
C=4π( R
i.e., C α R
As both the sphere have equal radii, they have the same capacitance. When charged to the same
potential, they will have the same charge.
10. (a) Given V=4x2 volt
i.e., electric potential changes only along X-axis.
Or
!)
!
Ex = - !* = - !* (4x2) = - 8x
electric field at point (1m, 0, 2 m) is
Ex = -8X1 = -8 Vm-1.
11. (a) The potential at any point inside the shell will be equal to that on its surface. When the shell is given a
charge -3Q the potential on its surface changes by same amount as that inside. Hence the potential
difference V between the surface of solid sphere and hollow sphere remains unchanged.
+
"
12. (a) V= =
,
- .
=-0.1 V.
13. (a) When the two conductors are connected by a wire, they share charges till their potential become equal.
.
'/
V1=V2
That is
"
. 0
'/ 0$
E1 =
Now,
10
1$
=
"0
.
"$
$$
0$
=
"0
0
=
And E2 =
0
$
.
$$
0$
=
$
0
"
. $
'/ $
"
. $
'/ $$
=
VA – VB =
or
"
[
2
4"
+
0
$
R
+q
]
$
52 $ 6!
"
"
[− 2 + $ $]
52 6!
"
"
[ − $ $
'/ 2
52 6!
=
= 2:1
14. (b) The two charged rings are shown in the figure.
VA =
'/
VB =
'/
"0
"$
Or
R
A
+
"
2
−
"
52 $ 6! $
]
=
"
/
d
[2 −
52 $ 6!$
-q
B
]
15. (c) The situation is shown in the figure.
Y
R1=OA=8(:2 − 0) + (:2 − 0) =2
VA =
"
.
'/ 0
VB =
"
.
'/ $
!)
!*
-=> ?-=@
'/
=
* $=A
16. (a) Given V (x)=
E= -
=
-=> ?-=@
.
'/
C
17. (c) V= . = Q. r1
VA – VB =0.
X
r
volts
'-*
= -20 (x2 – 4)-2 X2x= (* $ 4')
At x= 4 µm,
r2=OB=2-0=2
'-?'
-
E=(4') = '' =
V/µm.
B
!
D
Thus the potential on a conductor depends on the amount of charge, geometry and size of the conductor.
18. (a) V=
6"4"6"4"
[
]=
'/
5
0.
19. (b) The potential at every point of the circle will be same.
W= q ∆ V =qX0 =0.
20. (a) K.E. of the proton=Work done on the proton
=q ∆ V =eX1 kV =1 keV.
21. (c) Initial P.E. of the three charges
Ui=
'/
.
"0 "$ 6"$ "> 6"0 ">
Final P.E. of the three charges,
U f =9X109
??-=0$ 6?%?-=0$ 6?%?-=0$
]
=9X109[
??-=0$ 6?%?-=0$ 6?%?-=0$
-.G
W=U f – U I =(198-99)X10-13 = 99X10-3 J
BB?-=>
-.G
=
= 198X10-3 J
= 0.99 J ~ 0.01 J.
= 99X10-3 J
22. (a) ∆ U =U f – U I
">
'/
"0 "$
-.%
k = '/ [
+
"$ ">
-.
+
"0 ">
]
-.'
"0 "$
-.%
- '/ [
+
"$ ">
-.G
+
"0 ">
]
-.'
q3k = q2q3[-. − -.G]= 8 q2q3 k=8q2.
23. (a) Electric field is always perpendicular to the equipotential surface at any point.
V A = V B =V C.
24. (d) Potential is same at every point of the conducting sphere.
25. (c) Potential of big drop
=n2/3X Potential of a small drop
2.5= (125)2/3X V =25 V
V=0.1 V.
26. (c) Work done is zero. Force on the charge Q is along the radius and the direction of motion is perpendicular to
it.
W = F s cos900=0.
"
B?-H ?G- -=0$
"I
=
27. (b) V= '/ . =
28. (a) P=
+
=
'?'?-@
-.
=
W
B?-H ?G- ?.?-=0H
-=0$
= 7.2X104 V.
=b 160 MW.
29. (d) A positively charged body can have +ve, -ve or zero potential.
30. (b) Earth is a conducting sphere of large capacitance.
As C is very large, so V
V=
0 for all finite charges.
B
.
31. (a) The work done against the force of repulsion in moving the two charges closer increases the potential
energy of the system.
32. (a) Planes perpendicular to X-direction i.e., planes parallel to YZ- plane will be the equipotential surfaces.
33. (c) Half diagonal of the tube,
r=
5J$ 6J$ 6J$
=
5%J
P.E. of the charge +q at the center due to eight charges (each= - q) at the corner of the cube is
U = 8X '/
34. (c) VC =
"?(4")
"
[
'/ D.
+
4"
]
L.
=
= - 8X
"
[
/ M
"$
'/ 5%J/
"
M
− ]=0
=VD =
'"$
.
%/
5
J
"
[
'/ DN
+
4"
]
LN
Work done in moving the charge +Q along the semicircle CRD is
W = + Q(V D – V C) = +
"
Q(/ M
-0) =
"C
.
/ M
=
"
[
'/ %M
"
M
− ]=-
R
s
A
C
B
D
"
/ M