CBSE Class 12 · 2020 · ElectrostaticsLong Answer

(a) Using Gauss law, derive the expression for the electric field due to a spherical shell of uniform charge distribution

σ

and radius

R

at a point lying at a distance

x

from the centre of the shell, such that (i)

0 < x < R

, and (ii)

x > R

. (b) An electric field is uniform and acts along

+ x

direction in the region of positive

x

. It is also uniform with the same magnitude but acts in

- x

direction in the region of negative

x

. The value of the field is

E = 200 N/C

for

x > 0

and

E = - 200 N/C

for

x < 0

. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one flat face is at

x = + 10 cm

and the other is at

x = - 10 cm

. Find (i) the net outward flux through the cylinder, (ii) the net charge present inside the cylinder.

Solution

(a) Concentric Gaussian sphere of radius

x

.
(i) For

x < R

the enclosed charge is zero:

E = 0

(ii) For

x > R

the enclosed charge is

q = σ (4 π R^{2})

E (4 π x^{2}) = \frac{σ 4 π R ^{2}}{ε _{0}} \Rightarrow E = \frac{σ R ^{2}}{ε _{0} x ^{2}}

(b)(i) Flux leaves through both flat faces (the curved surface has

E

parallel to it):

ϕ = 2 E A = 2 \times 200 \times π (0.05)^{2} = 3.14 N m^{2} C^{- 1}

(ii) Net charge inside:

q = ε_{0} ϕ = 8.854 \times 1 0^{- 12} \times 3.14 = 2.78 \times 1 0^{- 11} C

CBSE Class 12 Board · 2020 · Subjective