1. Xtra Edge February 2012 - Career Point

Sol. (i) The capacitor A with dielectric slab can be
considered as two capacitors in parallel, one
having dielectric slab and one not having dielectric
A
slab each capacitor has an area of . The
2
combined capacitance is
110 V
A
C = C1 + C2
( A / 2)
ε 0 ( A / 2)
ε r
= +
0ε
=
d
d
=
0.
4
– 12
× 8.
85×
10
– 4
2×
8.
85×
10
+
+ + + +
– –
A/2
–
A/2
–
+
A ε 0
[1 + εr]
2
d
[1 + 9] = 2 × 10 –9 F
∴ Energy stored
1 2 1 –9 2 –5
= CV = × 2 × 10 × (110) = 1.21 × 10 J
2 2
(ii) Work done in removing the dielectric state =
(Energy stored in capacitor without dielectric) –
(Energy stored in capacitor with dielectric).
It may be noted that while taking out the dielectric
the charge on the capacitor plate remains the same.
∴ W =
q
2C'
2
q
–
2C
2
Here, C = 2 × 10 –9 F,
– 14
Aε 0 0.
04×
8.
85×
10
C' = =
= 0.4 × 10
– 4
d 8.
85×
10
–9 F
q = CV = 2 × 10 –9 × 110 = 2.2 × 10 –7 C
– 7
2
( 2.
2×
10 ) ⎡ 1 1 ⎤
∴ W = ⎢ –
– 9 – 9
2
⎥
⎣0.
4×
10 2×
10 ⎦
– 14
2.
2×
2.
2×
10 ⎡2
– 0.
4⎤
=
– 9 ⎢ ⎥
2×
10 ⎣ 2×
0.
4 ⎦
= 1.21 ×
1.
6
0.
4
× 10 –5 = 4.84 × 10 –5 J
ε 0ε r A B
(iii) The capacitance of B =
d
– 12
8.
85×
10 × 9×
0.
02
=
– 4
8.
85×
10
CB = 1.8 × 10 –9 F
The charge on A, qA = 2.2 × 10 –7 C gets distributed
into two parts.
∴ q1 + q2 = 2.2 × 10 –7 C Also the potential
difference across A = p.d. across B
XtraEdge for IIT-JEE 7 FEBRUARY 2012
∴
q 1 =
C
A
⇒ q1 =
q
C
B
2
B
C A . q2 =
C
0.
4×
10
1.
8×
10
∴ 0.22 q2 + q2 = 2.2 × 10 –7
2.
2
⇒ q2 =
1.
22
⇒ q1 = 0.4 × 10 –7 C
Total energy stored
=
2
1
A
q
2C
+
– 9
– 9
× 10 –7 = 1.8 × 10 –7 C
2
2
q
2C
B
– 14
. q2 = 0.22 q2
– 14
0.
4×
0.
4×
10 1.
8×
1.
8×
10
=
+
– 9
– 8
2×
0.
4×
10 2×
1.
8×
10
= 0.2 × 10 –5 + 0.9 × 10 –5 = 1.1 × 10 –5 J.
3. A particles of mass m = 1.6 × 10 –27 kg and charge q =
1.6 × 10 –19 C enters a region of uniform magnetic
field of strength 1 Tesla along the direction shown in
figure. The speed of the particle is 10 7 m/s. (i) the
magnetic field is directed along the inward normal to
the plane of the paper. The particle leaves the region
of the field at the point F. Find the distance EF and
the angle θ. (ii) If the direction of the field is along
the outward normal to the plane of the paper, find the
time spent by the particle in the region of the
magnetic field after entering it at E. [IIT-1984]
× × × × ×
θ × × × × ×
× × × × ×
F × × × × ×
× × × × ×
E × × × × ×
× × × × ×
× × × × ×
× × × × ×
45º
Sol. (a) m = 1.6 × 10 –27 kg, q = 1.6 × 10 –19 C
B = 1 T, v = 10 7 m/s
F = q . v B sin α
(acting towards O by Fleming's left hand rule)
F = qvB [Q α = 90º]
But F = ma
∴ qvB = ma
qvB
∴ a =
m
– 19
1.
6×
10 × 10 × 1
=
– 27
1.
6×
10
= 10 15 m/s 2
7