Page 17 - English Class X_cbse new (FINAL).cdr

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

18. By symmetry the electric field strength at every point on the flat 19. Charge enclosed by the cylindrical surface = λl [1]
Cont. surface is the same and its direction is normal outwards at the points [1]
By Gauss Theorem, electric flux = q
on the two plane surfaces and parallel to the curved surface. ε0 [2]

Total electric flux [1]

∫ ∫ ∫ ∫E • dS = E • dS1 + E • dS2 + E • dS3S3 φ = 1 × charge enclosed= 1 (λl) , φ = 1 (λl)
S S1 S2
ε0 ε0 ∈0

or [1]

E • dS = E + E dS3 cos90

∫ ∫ ∫ ∫S S1 S3
dS1 cos0 E dS2 cos0 +

S2

= E∫ dS1 + E∫ dS2 = Ea + Ea = 2Ea 20. Total charge within a surface S = +2q + (−q) = +q
∴ Electric flux φ = q
∴ Total electric flux = 2E a ε0

As σ is charge per unit area of sheet and a is the intersecting area,
the charge enclosed by Gaussian surface = σa

According to Gauss’s theorem,

Total electric flux = 1 × (totalcharge enclosedby the surface) [1]
ε0 [1]
21. Electric field due to a uniformly charged infinite plane sheet: Suppose
i.e., 2Ea = 1 (σa) ∴E = σ . a thin non-conducting uniform surface charge density σ.
ε0 2ε0
Electric field intensity E on either side of the sheet must be
Thus electric field strength due to an infinite slat sheet of charge is perpendicular to the plane of sheet having same magnitude to all
independent of the distance of the point and is directed normally points equidistant from sheet.
away from the charge. If the surface charge density σ is negative the
electric field is directed towards the surface charge. Let P be any point at a distance r from the sheet. Let the small area
element dS = dSn .
(c) (i) Away from the charged sheet.

(ii) Towards the plane sheet.

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