Page 12 - English Class X_cbse new (FINAL).cdr

S. Solution Marking S. Solution Marking
No. Scheme No. Scheme

2. parallel to the curved surface of the cylinder, the flux due to elec- 3. According to Gauss’s law [1]
Cont. tric field of the plane sheet of charge passes only through the two
[1] ∫φ = εo , ds = qenclosed
circular caps of the cylinder. If E is the magnitude of electric field at ∈0
the two circular caps of the cylinder. If E is the magnitude of elec-
tric field at point P, then electric flux crossing through the Gaussian Flux depends only on the charge enclosed. Hence, the electric flux
surface, Φ = E x area of the end faces(circular caps) of the cylinder remains constant.

Or ϕ = E x 2A (1)

According to Gauss theorem, we have ϕ = q/ε0 4. Umtormly charged
Here, the charge enclosed by the Gaussian surface, q =σA S2 intinite plane sheet

n

ϕ = σA /ε0 (2) E’
n
From equation (1) and (2),we have [1] n E
E x 2A = σA /ε0 E P
Or E = σ/2ε0 A
A’
∆S
∆S Cylindrical Gaussian

surface

Thus, we find that the magnitude of the electric field at a point due to
an infinite plane sheet of charge is independent of its distance from Consider a thin infinite uniformly charged plane sheet having the
the sheet of charge. surface charge density of σ. The electric field is normally outward to
the plane sheet and is same in magnitude but opposite in direction.
(b) Let V0 be the potential on the surface of sheet at a distance r Now, draw a Gaussian surface in the form of cylinder around on axis.
from it Let its cross-sectional area be A’ and the electric flux linked with S2 is
0. So, the total electric flux linked through the Gaussian surface is

dV = Edr ⇒ Vo − V = σ r ⇒ V = Vo − σ r φE = electric flux through A
2εo 2εo + electric flux throughS2
+ electric flux through A′

[1] φE = EA cos0° + 0 + EA cos0°. [1]
φ = 2EA …(i)

[5]

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