Page 10 - English Class X_cbse new (FINAL).cdr

Gauss Theorem

Sub Topic: 1.3

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

1. (a) Electric flux is defined as the number of electric field lines 1. at the centre of a cube of side 2d. Again the flux through [1]
crossing per unit area. Cont. one face of the cube will be 1/6 of the total electric flux [5]
due to the charge q.
It is given as ∆φ = E.∆Scosθ

Where E is electric field and ∆S is area vector. The Hence, the electric flux through the square will not
angle q is the angle between E and ∆S . Since electric flux is dot
product of two vectors. Hence it is a scalar quantity. [1] change and it will remain the same i.e. q
6ε0

OR

Field due to an infinitely long straight uniformly charged wire:

If we imagine a cube of side d, then this point charge q will be at the [1]
centre of the cube. [1]
[1]
Now, the total electric flux due to this charge will pass evenly through
the six faces of the cube.
Consider a thin, infinitely long straight line charge of linear density
So, the electric flux through one face will be equal to 1/6 of the total l. Let P be the point at a distance a from the line. To find the electric [1]
electric flux due to this charge. field at point P, draw a cylindrical surface of radius 'a' and length [1]
l. If E is the magnitude of electric field at point P, then electric flux
Now, the total electric flux generated by the charge will be given by through the Gaussian surface.
Gauss's law as:
f = E × Area of the curved surface of a cylinder of radius a and
∆φ = E.∆S = q length l . As electric lines of force are parallel to end faces (circular
ε0 caps) of the cylinder, there is no component of field along the normal
to the end faces.
q
So, the electric flux through the square = 6ε0 φ = E × A ⇒ φ = E × 2π a l − − − (1)

(b) If we move the charge to a distance d and the square According to Gauss's theorem,
side changes to 2d, still the point charge can be imagined

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