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CBSE Class 12 Physics 2018 Delhi Set 1 Paper

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Question : 24
Total: 26
SECTION - E

(a) Define electric flux. Is it a scalar or a vector quantity?
A point charge q is at a distance of d∕2 directly above the centre of a square of side d , as shown in the figure. Use Gauss' law to obtain the expression for the electric flux through the square.

(b) If the point charge is now moved to a distance ' d ' from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.
OR
(a) Use Gauss' law to derive the expression for the electric field (
→
E
) due to a straight uniformly charged infinite line of charge density λC∕m.
(b) Draw a graph to show the variation of E with perpendicular distance r from line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2>r1).
Solution:  
(a) Definition of electric flux
Stating scalar\/ vector
Gauss's Theorem
Derivation of the expression for electric flux
(b) Explanation of change in electric flux
(a) Electric flux through a given surface is defined as the dot product of electric field and area vector over that surface.
Alternatively ϕ=
≡
S
→
E
⋅
→
dS
‌
Also accept
Electric flux, through a surface equals the surface integral of the electric field over that surface.
It is a scalar quantity

Constructing a cube of side ' d ' so that charge ' q ' gets placed within of this cube (Gaussian surface) According to Gauss's law the Electric flux
ϕ‌=‌
‌ Charge enclosed ‌
ε0
‌=‌
q
ε0
This is the total flux through all the six faces of the cube.
Hence electric flux through the square
‌
1
6
×‌
q
ε0
=‌
q
6ε0
(b) If the charge is moved to a distance d and the side of the square is doubled the cube will be constructed to have a side 2d but the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore theflux through the square will remain the same as before.
OR
(a) Derivation of the expression for electric field
→
E
(b) Graph to show the required variation of the electric field
(c) Calculation of work done
(a)

To calculate the electric field, imagine a cylindrical Gaussian surface, since the field is everywhere radial, flux through two ends of the cylindrical Gaussian surface is zero.
At cylindrical part of the surface electric field
→
E
 is normal to the surface at every point and its magnitude is constant. Therefore flux through the Gaussian surface.
= Flux through the curved cylindrical part of the surface.
=E×2πrl .......(i)
Applying Gauss's Law
‌ Flux ‌ϕ=‌
q‌enclosed ‌
ε0
Total charge enclosed = Linear charge density ×I
‌=λl
ϕ‌=‌
λL
ε0
.......(ii)
Using Equations (i) & (ii)
E×2πrl‌=‌
λl
ε0
∴‌‌E‌=‌
λ
2πε0r
In vector notation
→
E
=‌
λ
2πε0r
^
n

(where
^
n
 is a unit vector normal to the line charge)
(b) The required graph is as shown:

(c) Work done in moving the charge ' q ' through a small displacement 'dr'
dW=
→
F
⋅
→
dr
dW‌=q
→
E
⋅
→
dr
‌=qEdr‌cos‌0∘
dW‌=q×‌
λ
2πε0r
dr
Work done in moving the given charge from r1 to
r2(r2>r1)
W=‌
r2
≡
r1
dW=
r2
∫
r1
‌
λqdr
2πε0r
W=‌
rq
2πε0
[loger2−loger1]
W=‌
λq
2πε0
[loge‌
r2
r1
]
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