Three charges each equal to 4 n C are placed at the three corners of a square of side 2 . 0 c m . Find electric field at the fourth corner.

Given q = 4 n C , a = 2 c m E 1 = Electric field due to charge q at distance a = ( 1 4 π ∈ 0 ) ( q a 2 ) E 2 = Electric field due to charge at distance 2 a = ( 1 4 π ∈ 0 ) q ( 2 a ) 2 = E 1 2 Net electric field at p, E n e t = 2 E 1 + E 2 2 (In the direction of E 2 ) = 2 E 1 + E 1 2 = ( 2 + 1 2 ) E 1 = ( 2 2 + 1 ) q 8 π ∈ 0 a 2 Putting values in the equation: = ( 2 2 + 1 ) 4 × 10 - 9 8 π ∈ 0 ( 2 × 10 - 2 ) 2 = ( 2 2 + 1 ) 10 - 5 8 π ∈ 0 V m - 1

State Gauss's law. Derive an expression for electric intensity at a point outside the uniformly charged shell.

Gauss's theorem in electrostatics: The total electric flux ϕ through any closed surface S in free space is equal to 1 ε ∘ times the total electric charge q enclosed by the surface. That is, ϕ = ∮ S E → . d s → = q ε ∘ , here ε ∘ is permittivity of free space. Electric field due to uniformly charged thin spherical shell at a point outside the shell: Here, O is the centre of the spherical shell of radius R , and charge enclosed by the spherical shell is q with uniformed charge density σ . Assuming the dotted circle of radius r to be Gaussian surface. Let a point P is outside the spherical shell. The distance between the centre of the spherical shell and the point P is r . Here, r > R . Now according to Gauss' theorem: ϕ = ∮ S E → . d s → = q ε ∘ ⇒ ϕ = ∮ S E d s cos θ = q ε ∘ Here, θ = 0 ° as E → and d s → are along the same direction. Thus, ϕ = ∮ S E d s = q ε ∘ ⇒ E ∮ S d s = q ε ∘ Since, ∮ S d s = 4 π r 2 Substituting this in above equation. E 4 π r 2 = q ε ∘ ⇒ E = 1 4 π ε ∘ q r 2 Hence, the expression for electric field at a point outside the uniformly charged thin spherical shell is E = 1 4 π ε ∘ q r 2 .

State Coulomb’s Law.

Coulomb's Law: The force between two electric charges at rest is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. If q 1 and q 2 are two charges having a distance r between them, then the magnitude of force between these two charges, F ∝ q 1 q 2 r 2 . F = K q 1 q 2 r 2 , where K is the constant. In SI units K = 1 4 π ε ο Nm 2 C 2 .

Give three properties of electric charge.

The properties of electric charges are as follows: (i) Like charges repel and unlike charges attract each other. (ii) Electric charges are quantized. (iii) Electric charges are additive. (iv) Electric charges are conserved. (v) The magnitude of elementary negative charge is same as that of elementary positive charge and is equal to 1 .6 × 10 − 19 C . (vi) Unlike mass, the electric charge on a body is not affected by its motion.

Write the SI unit of electric flux.

The SI unit of electric flux is N C - 1 m 2 . Electric flux is a scalar quantity. The electric flux ∆ φ through an area element ∆ S is defined by: ∆ φ = E → . ∆ S → = E ∆ S cos θ where θ is the angle between the electric field and the area vector.

Name two basic properties of electric charge.

There are very two basic properties of electric charge. They are: 1. Conservative nature of charge Electric charge of a particle is conservative in nature. It means that the charge can neither be created nor be destroyed 2. Quantization of charge The charge of a system is a fixed quantity. Technically, the charge is a quantized quantity. The net charge of a system can be expressed as the integral multiples of the basic unit of charge

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Karnataka Board-2020>Q 7

EASY

12th Karnataka Board

IMPORTANT

Earn 100

State and explain Coulomb's law in electrostatics.

Important Questions on Electric Charges and Fields

EASY

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Karnataka Board-2020>Q 8

Two point charges $q_{A} = 5 μ C and q_{B} = 5 μ C$ are located at A and B separated by $0.2 m$ in vacuum. What is the electric field at the midpoint O of the line joining the charges? Take $\frac{1}{4 π ε_{o}} = 9 \times 10^{9} {Nm}^{2} C^{- 2}$ .

EASY

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Karnataka Board-2020>Q 9

Two point charges $q_{A} = 5 μ C and q_{B} = 5 μ C$ are located at A and B separated by $0.2 m$ in vacuum. If a negative test charge $2 n C$ is placed at the midpoint O, what is the force experienced by the test charge?

Learn and Prepare for any exam you want !

|

|

|

|

|

|

|

|