There is an electric dipole in X-Y plane .Its dipole moment is 4 × 10 - 9 Cm On the same plane there is also a uniform electric field of magnitude 5 × 10 4 NC - 1 . If the axis of dipole makes an angle 30 ° with the electric field . Calculate the magnitude of torque acting on the dipole and also mention the direction of the torque

Given the electric dipole moment, p = 4 × 10 - 9 Cm . Angle made by p with a uniform electric field, θ = 30 ° . Electric field, E = 5 × 10 4 NC - 1 . The torque acting on the dipole, τ = pE sinθ = 4 × 10 - 9 × 5 × 10 4 sin 30 = 20 × 10 - 9 × 10 4 × 1 2 = 10 - 4 Nm . Hence, the torque is 10 - 4 Nm ., As the torque is the cross product of the field and the dipole moment, its direction is given by Fleming's right-hand rule. .

There is an electric dipole in X-Y plane .Its dipole moment is 4 × 10 - 9 Cm On the same plane there is also a uniform electric field of magnitude 5 × 10 4 NC - 1 . If the axis of dipole makes an angle 30 ° with the electric field . Calculate the magnitude of torque acting on the dipole and also mention the direction of the torque

Given the electric dipole moment, p = 4 × 10 - 9 Cm . Angle made by p with a uniform electric field, θ = 30 ° . Electric field, E = 5 × 10 4 NC - 1 . The torque acting on the dipole, τ = pE sinθ = 4 × 10 - 9 × 5 × 10 4 sin 30 = 20 × 10 - 9 × 10 4 × 1 2 = 10 - 4 Nm . Hence, the torque is 10 - 4 Nm ., As the torque is the cross product of the field and the dipole moment, its direction is given by Fleming's right-hand rule. .

State Gauss's law in electrostatics and write in its mathematical form.Calculate the electric field E → at point P due to charges this spherical shell, shown in figure (1) What will be the field E → shown in fig (2) and (3)? Given the surface charge density is σ .

Gauss Law for electrostatics states that the total electric flux passing through a closed surface equals the enclosed charge in the surface divided by the permittivity of the medium. Mathematically it is written as: ∮ Eds = q ε 0 Electric field due to a uniformly charged thin shell: In the diagram, O is the centre of the spherical shell of radius R , and charge enclosed by the spherical shell is q . Let a point P is outside the spherical shell, where we have to calculate the electric field. The distance between the centre of the spherical shell and the point A is r . The total charge on the Gaussian surface of radius R is, q = σ surface area = σ 4 πR 2 . Hence ∮ Eds = q 4 π R 2 σ ε 0 or E ∮ ds = q 4 π R 2 σ ε 0 . Case (1) If the point P lies outside the surface of the shell at a distance r from the centre of the sphere As ∮ E dS = ∮ dS = E . 4 πr 2 we have E ( 4 πr 2 ) = q 4 πR 2 σ ε 0 or E = σ ε 0 . R 2 r 2 ( for r > R ) Case (2) In figure (2) if the point lies at the surface of the sphere i.e R = r , so we have, E = σ ε 0 (Maximum) Case (3), In figure (3) The point lies inside the spherical shell. There is no charge inside a spherical shell hence the field is, E = 0

State Gauss's law in electrostatics and write in its mathematical form.Calculate the electric field E → at point P due to charges this spherical shell, shown in figure (1) What will be the field E → shown in fig (2) and (3)? Given the surface charge density is σ .

Gauss Law for electrostatics states that the total electric flux passing through a closed surface equals the enclosed charge in the surface divided by the permittivity of the medium. Mathematically it is written as: ∮ Eds = q ε 0 Electric field due to a uniformly charged thin shell: In the diagram, O is the centre of the spherical shell of radius R , and charge enclosed by the spherical shell is q . Let a point P is outside the spherical shell, where we have to calculate the electric field. The distance between the centre of the spherical shell and the point A is r . The total charge on the Gaussian surface of radius R is, q = σ surface area = σ 4 πR 2 . Hence ∮ Eds = q 4 π R 2 σ ε 0 or E ∮ ds = q 4 π R 2 σ ε 0 . Case (1) If the point P lies outside the surface of the shell at a distance r from the centre of the sphere As ∮ E dS = ∮ dS = E . 4 πr 2 we have E ( 4 πr 2 ) = q 4 πR 2 σ ε 0 or E = σ ε 0 . R 2 r 2 ( for r > R ) Case (2) In figure (2) if the point lies at the surface of the sphere i.e R = r , so we have, E = σ ε 0 (Maximum) Case (3), In figure (3) The point lies inside the spherical shell. There is no charge inside a spherical shell hence the field is, E = 0

You know, if a charge particle q, moving with velocity v → enters a uniform magnetic field B → , it experiences a force F → = q v × B → . Name the path described by q when the angle between v → and B → is 1 ) ∠ θ = 90 ° and 2 ) ∠ θ < 90 °

In terms of angle θ between the v → and B → , the magnitude of lorentz force F F → = q v × B → sin θ . 1 If the v → is perpendicular to B → i.e. θ = 90 ° The particle will follow a circular trajectory with a radius of r = m v → q B → m 2 If the v → is not perpendicular to B → i.e. θ < 90 ° The particle orbit will be helix with an axis parallel to lines.

Define one coulomb charge. Two point charges at a distance r in air exert a force F on each other. At what distance will these charges experience the same force F in a medium of dielectric constant k ?

By Coulomb' law, we have , F = 1 4 πε 0 . q 1 q 2 r 2 . Putting q = q ÷ and r = 1 m so F = 1 4 πε 0 = 9 × 10 9 N 1 coulomb is that quantity of electric charge which exerts a force of 9 × 10 9 N on an equal charge placed at a distance of 1 m from it. Given in air, we have the force, F = 1 4 πε 0 . q 1 q 2 r 2 , where r is the distance between the charge and ε 0 = permittivity of free space. If the dielectric constant of the medium is k , then the permittivity of the medium will be kε 0 and the distance between the charges for the same force be r ' then we have for the medium, F = 1 4 π k ε 0 . q 1 q 2 r ' 2 . As both are equal, we get, F = 1 4 πε 0 . q 1 q 2 r 2 = 1 4 πkε 0 . q 1 q 2 r ' 2 or 1 r 2 = 1 kr ' 2 Or r'= r k . This is the required expression.

Mention one similarity between Coulomb forces and gravitational forces acting between two stationary charges.

One similarity between coulomb force and gravitational force is that both coulomb force and the gravitational force is inversely proportional to the distance between charges and masses respectively. Both forces obey inverse square law

Derive the expression for field intensity due to an electric dipole in vacuum for points on its axis. Compare the variation of field intensity with distance for an electric dipole and a point charge when both are kept in a vacuum.

Consider an electric dipole of two-point charge + q and - q separated by a small distance 2 l .Consider a point P on its axial line at a distance r from its centre O . The electric field intensity at P due to charge at A , E → 1 = 1 4 πε 0 . q r + a 2 along PA ...............................................(1) The electric field intensity at P due to charge at B , E → 2 = 1 4 πε 0 . q r - a 2 along BP . The net electric field is given by, E = E 1 + E 2 = 1 4 πε 0 . q r + a 2 + 1 4 πε 0 . q r - a 2 = 1 4 πε 0 q r - a 2 + q r + a 2 along BP . = q 4 πε 0 r + a 2 - r - a 2 r 2 - a 2 2 = q r - a 2 4 ar r 2 - a 2 2 = P 2 r r 2 - a 2 2 . Where 2 ql = dipole moment. If the dipole is short, 2 a < < l , then we get, E → = 2 r 4 πε 0 . r 4 = 2 P 4 πε 0 r 3 ..............(1) direction along BP . It is clear that E ∝ 1 r 3

Apply Gauss's law to derive the expression for electric field intensity due to an infinitely long straight uniformly charged wire. What is the direction of the field intensity it is positively charged?

According to Gauss's law, the total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. Gauss's law is true for any closed surface, no matter what its shape or size. ∫ s E → d S → cosθ = q ε 0 Consider an infinitely long thin straight wire with uniform linear charge q and density λ . Let P be a point at distance r from the wire. Draw a cylindrical surface of radius r with point P on its surface. The cylindrical surface is a Gaussian surface. The total charge enclosed by Gaussian surface due to continuous charge distribution along the length l of the wire is given by, q = λl . From Gauss's law, we have, ∫ s E d S → cosθ = q ε 0 = λl ε 0 or E ∫ s dS = λl ε 0 as for the curved surface, the angle between the direction of the field and the normal to the surface is zero. As the total surface area ∫ dS = 2 πrl , we get, E 2 πrl = λl ε 0 or E = 1 2 πε 0 . λ r . The direction of the field intensity will be vertically outwards of the surface of the wire.

Two-point charges 0 . 01 μC and - 0 . 01 μC are placed 10 cm in vacuum. Calculate the magnitude of the electric field intensity at the middle point of the line joining the charges and mention its direction.( K = 9 × 10 9 NM 2 C 2 )

Given two charges A = 0 . 01 μC and B = - 0 . 01 μC . The separation between them= AB = 10 cm .. To find the electric field at their midpoint O such that A - B = 5 cm . The field at O due to A , E 1 = k × 0 . 01 × 10 - 6 5 × 10 - 2 NC - 1 along AO , where k = 1 4 πε 0 = 9 × 10 9 NM 2 C 2 The field on O due to B = E 2 = k × 0 . 01 × 10 - 6 5 × 10 - 2 NC - 1 along OB . Therefore, the total field E = E 1 + E 2 = k × 0 . 01 × 10 - 6 5 × 10 - 2 NC - 1 + k × 0 . 01 × 10 - 6 5 × 10 - 2 NC - 1 E = 9 × 10 9 2 × 0 . 01 × 10 - 6 5 × 10 - 2 2 = 0 . 072 × 10 7 = 72 × 10 3 NC - 1 The direction along negative charge side.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>assam board-2019>Q 8

EASY

12th Assam Board

IMPORTANT

Earn 100

There is an electric dipole in X-Y plane .Its dipole moment is $4 \times 10^{- 9} Cm$ On the same plane there is also a uniform

electric field of magnitude $5 \times 10^{4} {NC}^{- 1}$ . If the axis of dipole makes an angle $30 °$ with the electric field . Calculate the magnitude of torque acting on the dipole and also mention the direction of the torque

Important Questions on Electric Charges and Fields

EASY

12th Assam Board

IMPORTANT

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

Mention one difference between mass and charge.

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>assam board-2020>Q 10

A closed spherical surface encloses a charge q at its centre. Show the electric flux through the closed surface is $q / ε_{\circ}$ .

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>assam board-2020>Q 12

State Gauss's law in electrostatics and write in its mathematical form.Calculate the electric field $\vec{E}$ at point $P$ due to charges this spherical shell, shown in figure (1) What will be the field $\vec{E}$ shown in fig (2) and (3)? Given the surface charge density is $σ$ .

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