Define polar and non-polar molecules. Explain the behaviour of non-polar molecules in a uniform magnetic field.

Nonpolar molecules: The molecule whose charge distribution is spherically symmetric when averaged over time; since the charges oscillate, a temporary dipole moment exists at any given instant in a so-called nonpolar molecule. The centre of mass of positive charges coincides with the centre of mass of the negative charges, in the molecule. The dipole moment of each molecule in the non- polar system is zero. Examples of homo nuclear nonpolar molecules are oxygen (O 2 ), nitrogen (N 2 ), and ozone (O 3 ). Other nonpolar molecules include carbon dioxide (CO 2 ) and the organic molecules' methane (CH 4 ), toluene, and gasoline. Most carbon compounds are nonpolar. Polar molecules: A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. A diatomic molecule that consists of a polar covalent bond, such as HF, is a polar molecule. The dipoles do not cancel out resulting in a net dipole. Examples: water, ammonia, sulphur dioxide These have the following characteristics: The centre of mass of positive charges does not coincide with the centre of mass of the negative charges, in the molecule. In its normal state (i.e., in the absence of any external force,(electric field), each molecule has some intrinsic permanent dipole moment and These have an asymmetric shape. Behaviour of non polar molecules in uniform magnetic field In an electric field, this force causes the electron cloud of a nonpolar molecule to be displaced in the direction of the attraction.

Write Coulomb's law and represent the forces between the two charges in vector form.

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects. In equation form, Coulomb's law can be stated as. F ∝ q 1 q 2 r 2 where q 1 , q 2 are the two charges and r is the distance between the two, or F = K q 1 q 2 r 2 where K is the constant of proportionality and represented by, K = 1 4 π ε 0 ε 0 is the permittivity of free space. In vector form it can be written as , F → 12 = 1 4 πε 0 q 1 q 2 r 12 2 × r ^ 21 = - F → 21 where the force on one due to other is opposite to each other. or we can write F → 12 = 1 4 πε 0 q 1 q 2 r 12 3 × r → 12

An electric dipole is placed in a uniform electric field. The resultant force on it is _______.

Depends upon relative position.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Gujarat Board-2019>Q 9

EASY

12th Gujarat Board

IMPORTANT

Earn 100

Two metallic spheres of radii $R_{1}$ and $R_{2}$ are charged. Now they are brought into contact with each other with a conducting wire and then are separated if the electric fields on their surfaces are $E_{1}$ and $E_{2}$ respectively, $\frac{E_{1}}{E_{2}} =$ ______.

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Important Questions on Electric Charges and Fields

EASY

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Gujarat Board-2019>Q 10

Define surface charge density and volume charge density of an electric charge. Also, state its unit.

MEDIUM

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Gujarat Board-2019>Q 11

A charge

Q

is uniformly distributed on the circumference of a circular ring of radius

a

. Find the intensity of the electric field at a point at a distance

x

from the centre of the axis of the ring.

MEDIUM

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Gujarat Board-2020>Q 16

An electron falls through a distance of $1.5 cm$ in a uniform electric field of magnitude $2 \times 10^{4} {NC}^{- 1}$ . The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance. Compute the time of fall in each case. (Given: $m_{e} = 9.11 \times 10^{- 31} kg, m_{p} = 1.67 \times 10^{- 27} kg,$ $q_{e} = - 1.6 \times 10^{- 19} C, q_{p} = 1.6 \times 10^{- 19} C$ )