Page 11 - English Class X_cbse new (FINAL).cdr

S. Solution Marking S. Solution Marking
No. Scheme No. Scheme
Let's suppose the charge is moved by an infinitely small dis-
1. φ= q 1. tance dr. Small work is done in moving the charge, dw = qdv
ε0 [1] Cont. where V is the potential difference between the two points.
Cont. [1]

∵ q = λl

∴ φ = λl − − (ii) Now, dV = −∫ E.dr
ε0
r2 λdr = λ r2 dr =λ  r1 
r1 2πrε0 2πε0 r1 r 2πε0  r2 
From equation (i) and (ii) we get; ∫ ∫= ln  

E × 2π al = λl Workdone = qdV = qλ ln  r1 
ε0 2πε0  r2 
 
E= 1 λ
2πε0 a [5]

(b) From the equation

E= λ 2. (a) Electric Field due to an infinite plane sheet of charge:-
2πε0r
Umtormly charged
S2 intinite plane sheet
where r is perpendicular distance from the line of charge, we can
n
see that E ∝ 1 [1]
r E’
n
n E
E P
A
A’
∆S
∆S Cylindrical Gaussian

surface

Consider an infinite thin plane sheet of positive charge having a [1]
(c) uniform surface charge density σ on both sides of the sheet. By
[1] symmetry, it follows that the electric field is perpendicular to the
plane sheet of charge and is directed in outward direction. To find
electric field due to the plane sheet of charge at any point P distant
r from it, draw a cylinder of area of cross section A through the point
P as the guassian surface. Since electric lines of force are parallel
to the curved surface of the cylinder, the electric lines of force are

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