A point charge of 4 . 427 μC is at the centre of a cubic Gaussian surface 9 . 0 cm on the edge. Find the net electric flux through the surface. ε 0 = 8 . 854 × 10 - 12 C 2 N - 1 m - 2

Net electric flux ϕ N e t through the cubic surface is given by, ϕ N e t = q ε 0 ε 0 = Permittivity of free space = 8.854 × 10 - 12 N - 1 C 2 m - 2 q = Net charge contained inside the cube = 2.0 μ C = 2 × 10 - 6 C ϕ N e t = 2 × 10 - 6 8.854 × 10 - 12 = 2.26 × 10 5 N m 2 C - 1 The net electric flux through the surface of cube is 2.26 × 10 5 N m 2 C - 1 .

What is an electric field? Derive an expression for electric field due to an electric dipole at any point along its equatorial line.

Electric field is a force produced by a charge near its surroundings. This force is exerted on other charges when brought in the vicinity of this field. Electric field due to a charge at a point is the force that a unit positive charge would experience if placed at that point. Electric field at A due to the charges q and - q Is shown in the above figure. We resolve E Into horizontal and vertical components. The vertical components ( E sinθ ) Cancel out each other and only the horizontal components survive to give net electric field at A as 2 Ecosθ . Electric field at point A , E A = 2 Ecosθ where E = q 4 πε ∘ r 2 + a 2 cos θ From figure , we have cos θ = a r 2 + a 2 ∴ E A = 2 qa 4 πε ∘ r 2 + a 2 3 / 2 We know dipole moment P = 2 qa ⇒ E A = P 4 πε ∘ r 2 + a 2 3 / 2 For a < < r , we can neglect a 2 compared to r 2 . ⇒ E A = P 4 πε ∘ r 3

Define an electric dipole.

The electric dipole is a pair of two equal and opposite charges, + q and - q Are separated by a very small distance. The algebraic sum of the charge on the dipole is zero and the electric field of the dipole is not zero, thus the electric field of dipole is added.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Manipur Board-2020>Q 5

EASY

12th Manipur Board

IMPORTANT

Earn 100

Define dipole moment vector of an electric dipole. Give the direction of the vector.

Important Questions on Electric Charges and Fields

HARD

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Manipur Board-2020>Q 6

Using Gauss's law of electrostatics obtain the expression for the electric field at a point near an infinitely long thin straight wire with uniform linear charge density

λ

.

Learn and Prepare for any exam you want !

|

|

|

|

|

|

|

|