Page 19 - English Class X_cbse new (FINAL).cdr

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

23. (i) Curved surface S1 (ii) Flat surface S2 and (iii) Flat surface S3. 24. Gauss Law: Total electric flux through a closed surface enclosing a
cont. By symmetry, the electric field has the same magnitude E at each 1
point of curved surface S1 and is directed radially outward.
[1] charge is equal to ε0 times the magnitude of the charge enclosed.

EE dS1 E Electric field Due to a uniformly charged infinite plane sheet:
EE Suppose a thin non-conducting infinite sheet of uniform surface,
charge density σ

90º S1 r 90º
dS2 dS3

S2 S3 n n

We consider small elements of surfaces S1, S2 and S3. The surface E P E
element vector dS1 is directed along the direction of electric field ds n ds

(i.e., angle between E and dS1 is zero); the elements dS2 and n n
dS3 are directed perpendicular to field vector E (i.e., angle between Q
dS2 and E is 90° and so also angle between dS3 and E ). n

Electric Flux through cylindrical surface r r

[1] Gaussian surface for a uniformly charged infinite plane sheet

∫ ∫ ∫ ∫Ei dS = Ei dS1 + Ei dS2 + Ei dS3 Electric field intensity E on either side of the sheet must be
S S1 S2 S3 perpendicular to the plane of sheet having same magnitude at all

points equidistant from sheet.

∫ ∫ ∫= S1 E dS1 cos 0 + S2 E dS2 cos 90 + S3E dS3 cos 90 Let P be any point at a distance r from the sheet. Let the small area
element ds = ds ⋅n .
∫= SE dS1 + 0 + 0

= E∫ dS1 (since electric field E is the same at each point of curved surface) E and n are perpendicular, on the surface of imagined cylinder,
so electric flux is zero. E and n are parallel on the two cylinderical
= E 2πrl (sinceareaof curvedsurface = 2πrl) [1]
edges P and Q, which contributes electric flux.
As λ is charge per unit length and length of cylinder is l, therefore, 1
charge enclosed by assumed surface = (λl) ∴Electric flux over the edges P and Q of the cylinder is  2

∴ By Gauss’s theorem 2φ = q ⇒ 2 ∫ E.ds = q ∵ φ = ∫ E.ds
∈0 ∈0

∫ EidS = 1 × charge enclosed 2∫ E.ds = q
ε0 ∈0

⇒ E.2πrl = 1 (λl) ⇒ E = λ [1] 2∫ E ds = q ∵E ⊥ ds
[3] ∈0
ε0 2πε0r
2Eπr2 = q
Thus, the electric field strength due to a line charge is inversely ∈0
proportional to r.

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