Test: Electric Flux & Electric Dipole (NCERT) - NEET MCQ

 As electric flux, φ = E.Δs
∴ unit of φ is N C^-1 m²

 Electric flux, φ= 
The dimension of φ = Dimension of E x dimension of s
[φ] = [M¹L¹T^-2][AT]^-1[L²] = [M¹L³T^-3A^-1]

Here, r = 10 cm 0.1 m, E = 5 x 10⁵ N C^-1
As the angle between the plane sheet and the electric field is 60⁰, angle made by the normal to the plane sheet and the electric field is θ = 90⁰ - 60⁰ = 30⁰
φ_E = ES cos θ = E x πr² cos θ
= 5 x 10⁵ x 3.14 x (0.1)² cos 30⁰ = 1.36 x 10⁴ N m²C^-1

Here, E = 2 x 10³ N C^-1 is along + x-axis
Surface area, s = (10 cm)² = 10² x 10^-4 m² = 10^-2 m²
When plane is parallel to yz plane, θ = 0°
So φ = E s cosθ = 2 x 10³ x 10^-2 cos 0^° = 20N C^-1 m²

When normal to the plane makes an angle of 60° with x-axis, then θ = 60°
φ = Es cosθ = 2 x 10³ x 10^-2 cos 60° = 10 N C^-1m²

Dipole moment is a vector quantity and has magnitude of 2qa and it is in the direction of the dipole axis from -q to q.

The unit of electric dipole moment is: 2ql =  C − m = Debye

Here, q = ±20μC = ±20 x 10^-6C, 2a = 10 mm = 10 x 10^-3m
r = OP = 10 cm = 10 x 10^-2 m
= q x 2a = 20 x 10^-6 x 10 x 10^-3 m = 2 x 10^-7 m
The electric field along BP, 
As a < < r,

When there are various types of charges in a region, but the total charge is zero, the region can be supposed to contain a number of electric dipoles. Therefore, at points outside the region, the dominant electric field ∝ (1/r³) for large r.

The Point lies on equatorial line of a short dipole,