Page 16 - English Class X_cbse new (FINAL).cdr

S. Solution Marking S. Solution Marking
No. Scheme No. Scheme
14. Same as answer 25 . .-2Q +Q
15. [1] 17. Electric flux through whole cube = Q . [1]
(a,0) (4a,0) ε0
(0,0)
Electric flux through one face = 1 Q µVm.
6 ε0

Electric flux, φ = −2Q [1] Explanation:
∈0 [2]
Net electric flux (φ) = q
Concept: Imagine a sphere of radius 3a about the origin and observe [1]
that only charge -2Q is inside the sphere. Use Gauss theorem to get [1] ∈0
the result. [2] Total faces of cube = 6
so flux from one cube = q

6 ∈0

16. (i) Electric flux through a surface, φ = E⋅S 18. (a) E lectric flux: It is defined as the total number of electric filed lines
passing through an area normal to its surface.
[1]
Flux through the left surface, φL = − E S = −50x. S Also, φ = ∫ E. ds [1]

Since x = 1m, φL = −50 × 1× 25 × 10−4 The SI unit is Nm2/C or volt-metre

= −1250 × 10−4 = −0.125 N m2 C−1 (b) L et electric charge be uniformly distributed over the surface of a
thin, non-conducting infinite sheet. Let the surface charge density
Flux through the right surface, (i.e., charge per unit surface area) be σ. We need to calculate
field strength at any point distance r from the sheet of charge.
φR = E S

Since x = 2m, φR = 50x S θ= 90º
dS3
= 50 × 2 × 25 × 10−4 = 2500 × 10−4 = 0.250 N m2C−1 S2
S1 E 90º S3
Net flux through the cylinder, φnet = φR + φL
dE S1 dS2
= 0.250 − 0.125 = 0.125Nm2 C−1 θ= 0º E
θ= 0º
(ii) Charge inside the cylinder, by Gauss’s Theorem B A

Sheet

φnet = q ⇒q= ε0 φNet To calculate the electric field strength near the sheet, we now
ε0 consider a cylindrical Gaussian surface bounded by two plane faces
A and B lying on the opposite sides and parallel to the charged sheet
= 8.854 × 10−12 × 0.125 and the cylindrical surface perpendicular to the sheet (fig).

= 8.854 × 10−12 × 1 = 1.107 × 10−12 C

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