Page 15 - English Class X_cbse new (FINAL).cdr

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

10. When the plane is parallel to the y – z plane [1] 12. Gauss’s theorem states that the electric flux through a closed surface

φ = E A here, E = 2 × 103 i [1] enclosing a charge is equal to 1 times the magnitude of the charge [1]
enclosed. ε0
[2]
E = 2 × 103 iN / CA

Side of square = 20 cm = 20 × 10−2 -Q + 4Q
Area of square (A) = 20 × 20 × 10−2 × 10−2 = 4 × 10−2 m2 (-3a, 0) (0, 0) (a, 0) (2a, 0)
∴ φ = 2 × 103 i × 4 × 10−2 i

⇒ φ = 80 weber

when the plane makes a 30° angle with the x-axis, the area vector The sphere enclose charge = + 4Q [1]
makes a 60° angle with the x-axis. [2]

φ = E A ⇒ φ = cosθ = 2 × 103 × 4 × 10−2 × cos 60 Thus, φ = 4Q
ε0
⇒ 2 × 103 × 4 × 10−2 cos 60 = 80 × 1
2 13. Gauss’s theorem states that the electric flux through a closed surface
enclosing a charge is equal to 1 times the magnitude of the charge
⇒ φ = 80 / 2 ⇒ φ = 40 weber ε0
enclosed. The sphere enclose charge = – 3Q + 2Q = –Q
11. Here, E = 5 × 103 iN / C , i.e. field is along positive direction of X-axis.
Surface area, A = 10 cm × 10 cm=0.10m × 0.10m = 10−2m2 [1] [1]
[1]
(i) When plane parallel to Y-Z plane, the normal to plane is along [2] ⇒ φ = −Q [1]
ε0 [2]
X-axis. Hence θ = 0

φ = EA cos θ = 5 × 103 × 10−2 cos 0 = 50NC−1m2

(ii) When the plane makes a 30° angle with the X-axis, the normal

to its plane makes 60° angle with X-axis. Hence θ = 60°

φ = EA cos θ = 5 × 103 × 10−2 cos 60 = 25NC−1m2

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