IIT-JEE 2011 - Career Point

Hence, E x = –
∴
∂V
∂ x
E = – ayi – axj
= – a[yi + xj]
∂V
= –ay, E y = –
∂ y
= – ax
3. A charge Q is distributed over two concentric hollow
spheres of radii r and R (> r) such the surface
densities are equal. Find the potential at the common
centre.
q′ q
O
R
Sol. Let q and q′ be the charges on inner and outer sphere.
Then
q + q′ = Q
…(1)
As the surface densities are equal, hence
q q'
=
2 2
4πr
4πR
(∴ Surface density = charge/area)
∴ q R 2 = q′ r 2
…(2)
From eq. (1) q′ = (Q – q), hence
q R 2 = (Q – q)r 2
q(R 2 + r 2 ) = Q r 2
2
2
Q r
Q R
∴ q = and q′ = Q – q =
2 2
2 2
R + r
R + r
Now potential at O is given by
1 q 1 q'
V = +
4πε0
r 4πε0
r
=
=
1
4πε
0
0
2
Q r 1
+
2 2
(R + r ) r 4πε
Q (r + R)
2
4πε (R r )
2 +
r
0
(R
Q r
2
2
2
+ r ) r
4. S 1 and S 2 are two parallel concentric spheres enclosing
charges q and 2q respectively as shown in fig.
(a) What is the ratio of electric flux through S 1 and S 2 ?
(b) How will the electric flux through the sphere S 1
change, if a medium of dielectric constant 5 is
introduced in the space inside S 1 in place of air ?
S 2
Sol. (a) Let Φ 1 and Φ 2 be the electric flux through spheres
S 1 and S 2 respectively.
q
2q
S 1
∴
q
Φ 1 =
ε
0
q / ε
Φ 1 =
0
Φ 2 3q / ε0
and Φ 2 =
= 3
1
q + 2q 3q
=
ε ε
(b) Let E be the electric field intensity on the surface of
sphere S 1 due to charge q placed inside the sphere.
When dielectric medium of dielectric constant K is
introduced inside sphere S 1 , then electric field
intensity E′ is given by
E′ = E/K
Now the flux Φ′ through S 1 becomes
' 1 q
Φ′ =
∫
E .dS =
∫
E.dS =
K Kε
∴ Φ′ =
q
5ε
0
5. A charge of 4 × 10 –8 C is distributed uniformly on the
surface of a sphere of radius 1 cm. It is covered by a
concentric, hollow conducting sphere of a radius
5 cm. (a) Find the electric field at a point 2 cm away
from the centre. (b) A charge of 6 × 10 –8 C is placed
on the hollow sphere. Find the surface charge density
on the outer surface of the hollow sphere.
Sol. (a) See fig. (a) Let P be a point where we have to
calculate the electric field. We draw a Gaussian
surface (shown dotted) through point P. The flux
through this surface is
q = 6 × 10 –8 C
5cm
Φ =
2cm
P
Fig. (a) Fig. (b)
∫
∫
E.dS = E dS = 4π(2×
10 ) E
0
0
0
−2
2
According to Gauss's law, Φ = q/ε 0
∴ 4π × (2 × 10 –2 ) 2 E = q/ε 0
9
−8
q (9×
10 ) × (4×
10 )
or E =
=
−2
2
−4
4πε0 × (2×
10 ) 4×
10
= 9 × 10 5 N/C
(b) See fig. (b) We draw a Gaussian surface (shown
dotted) through the material of hollow sphere. We
know that the electric field in a conducting material is
zero, therefore the flux through this Gaussian surface
is zero. Using Gauss's law, the total charge enclosed
must be zero. So, the charge on the inner surface of
hollow sphere is 6 × 10 –8 C. So, the charge on the
outer surface will be 10 × 10 –8 C.
XtraEdge for IIT-JEE 27 MAY 2010