Page 18 - English Class X_cbse new (FINAL).cdr

S. Solution Marking S. Solution Marking
No. Scheme No. Scheme
σ1 σ2
21. E and n are perpendicular on the surface of imagined cylinder, 21.

Cont. Cont.

so electric flux is zero. E and n are parallel on the two cylindrical
edges P and Q, which contributes electric flux.

E n n B
ds A
P ds
n [1]
E
n The electric field between two plates is given by
Q
n

E =E1 -E2

rr = σ1 − σ2 = σ −  −σ  = σ+σ [1]
2ε0 2ε0 2ε0  2ε0  2ε0 [3]
∴ Electric flux over the edges P and Q of the cylinder is
∴ E = 2σ ⇒ E = σ [1]
∫2φ = q = 2 E ⋅ dS = q 2ε0 ε0 [1]
[2]
ε0 ε0
16
∫⇒ 2 E dS = q ⇒ 2E πr2 = q ⇒ E = q 22. Electric flux: Electric flux over an area in an electric field represents
ε0 2πε0r 2 the total number of electric field lines crossing this area. It is denoted
ε0 by φE .

∴ The change density σ = q The SI unit of electric flux is NM2 C-2.
S

⇒ q = πr2σ The electric flux is given by

E = πr 2 σ =∫φEE.ds = q
2πε0r 2 ε0
S

E= σ ,vectoricallyE = σ n [1] which is independent of the radius of sphere. Hence the electric flux
2ε0 2ε0 through the surface of sphere remains same.

Where n is a unit vector normal to the plane and going away from it. 23. Electric field due to infinitely long, thin and uniformly charged
straight wire: Consider an infinitely long line charge having line-
Where σ > 0E is directed away from both sides. ar charge density λ coulomb metre-1 (linear charge density means
charge per unit length). To find the electric field strength at a distance
Now consider two infinite plane parallel sheets of charge A and B. Let r, we consider a cylindrical Gaussian surface of radius r and length
σ1 = + σ and σ2 = - σ be the uniform surface densities of charge on A I coaxial with line charge. The cylindrical Gaussian surface may be
and B respectively. divided into three parts: