The Electric Field Due to an Electric Dipole MCQ Quiz in मल्याळम - Objective Question with Answer for The Electric Field Due to an Electric Dipole - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT :

• The electric field for the points on the axis of the dipole is given by

\(E = \frac{2P}{4\pi\epsilon_{0}(r^{2}-a^{2})^{2}}\)

if r>>>>>a then the above expression becomes

\(E = \frac{2P}{4\pi\epsilon_{0}r^{3}}\)

Where P = Diploe momentum, r = Distance, a = Half of distance between both the charge.

• The electric field for the points on the equatorial plane is given by

​\(E = -\frac{P}{4\pi\epsilon_{0}(r^{2}+a^{2})^{\frac{3}{2}}}\)

• if the value of r >>>>a the electric field is given by

\(E = -\frac{P}{4\pi\epsilon_{0}r^{3}}\)

Where P = Diploe momentum, r = Distance, a = Half of distance between both the charge.

EXPLANATION : 

• The electric field along the axial point is given by

\(E = \frac{2P}{4\pi\epsilon_{0}r^{3}}\)       ----(1)

• The electric field along a point in the equatorial plane is given by

\(E = -\frac{P}{4\pi\epsilon_{0}r^{3}}\)        ----(2)

Comparing both equations 1 and 2 we get

\(\Rightarrow E\propto\frac{1}{r^{3}}\)

• The electric field is inversely proportional to the cube of the distance. Hence option 4 is the answer

The correct answer is Always attract each other.

• Electric charge is the physical property of matter that experiences a force when placed in an electromagnetic field.

Key Points

• There are two types of electric charges i.e. positive and negative.
• Like charges repel each other whereas unlike charges attract each other.
• Thus, two negative charges tend to repel one another, while a positive charge attracts a negative charge.
• An object with an absence of net charge is said to be neutral.
• The attraction or repulsion acts along the line between the two charges.

CONCEPT:

Electrostatic External Field Torque- It is defined as the product of the moment and field and it is written as;

Electrostatic External Field Energy - It is defined as the dot product of moment and the field is written as;

CALCULATION:

As we know;

Electrostatic External Field Torque- It is defined as the product of the moment and field and it is written as;

​\(\tau = p\times E\)​

Electrostatic External Field Energy - It is defined as the dot product of moment and the field is written as;

\(U=-p.E \)

The electrostatic axial field for a short dipole - The Electrostatic axial field for a short dipole is written as;

\(d=\frac{2p}{4\pi\epsilon_o r^3} \)

Electrostatic Equatorial Field for a short Dipole is written as;

\(d_E=\frac{-p}{4\pi\epsilon_o r^3}\)

Hence option 1) is the correct answer.

CONCEPT:

• Electric dipole: When two equal and opposite charges are separated by a small distance then this combination of charges is called an electric dipole.
  • The multiplication of charge and the distance between them is called an electric dipole moment.
  • The electric dipole moment is denoted by P and the SI unit of dipole moment is Coulombmeter (Cm)

Dipole moment (P) = q × d

Where q is charge and d is the distance between two charged particles.

• The direction electric dipole moment is from a negative charge to a positive charge.
• The space or region around the electric charge in which electrostatic force can be experienced by other charged particles is called an electric field by that electric charge.

Here d = 2a

The electric field on equatorial line (at point A):

E = -qa / [2πεo(r2 + a2)3 / 2]

When r >>> a

Then 

\(\overrightarrow{{{E}_{eq}}}=\frac{-~1}{4\pi {{\epsilon }_{0}}}\frac{~\vec{P}}{{{r}^{3}}}\)

Where vector P is the dipole moment of the electric dipole, ϵ0 is the permittivity of free space and r is the distance of points A and B from the centre of the dipole.

EXPLANATION:

The magnitude of electric field of a dipole at a point on the equatorial plane is, E = -qa / [2πεo(r2 + a2)3 / 2]

So option 4 is correct.

Gauss’s Law:

The total of the electric flux out of a closed surface is equal to the charge (Q) enclosed divided by the permittivity (ϵ0) of the medium. It can be given by ϕE

​\({ϕ _{\bf{E}}} = \frac{{\bf{Q}}}{{{\epsilon_0}}}\)​

Analysis:

A charge is placed at the center of the cube as shown:

Flux is linked equally with all the faces (6 faces)

∴ Using Gauss’s theorem, the flux linked with 1 face will be:

\({{ϕ }_{E}}=\frac{q}{{{\epsilon }_{0}}}\)

As a cube is given, flux linked with 6 faces are:

\(6× {{ϕ }_{E}}=\frac{q}{{{\epsilon }_{0}}}\)

\({{ϕ }_{E}}=\frac{q}{6~{{\epsilon }_{0}}}\)

Calculation:

With q = 5.0 μC, the electric flux through any one face of the cube will be:

​\({{ϕ }_{E}}=\frac{5× 10^{-6}}{6~{× 8.854× 10^{-12}}}\)​

ϕ_E = 9.4 × 10⁴ N m2/C

Electric displacement:

Electric displacement, denoted by D, is the charge per unit area that would be displaced across a layer of conductor placed across an electric field. It is also known as electric flux density.

Electric displacement is used in the dielectric material to find the response of the materials on the application of an electric field E.

In Maxwell’s equation, it appears as a vector field.

The SI unit of electric displacement is Coulomb per meter square (C m-2).

Mathematically relation is given by

P = D - ε0E

where, 

ϵ0 = Vacuum permittivity,

P = Polarization vector,

E = Electric field,

D = Electric displacement field

Derivation:

We know that the effect on dielectric placed in an external electric field E0 and an electric field due to polarized charges, this field is called electric field due to polarization (Ep) in relation with electric field vector E.

E = E0 – Ep    ---(1)

Polarization vector, P is equal to the bound charge per unit area or equal to the surface density of bound charges (because surface charge density is a charge per unit area),

Thus \(P = {q_b\over A} = σ_p\)   ---(2)

Where qb is bound to charge and σp is surface density of bound charges.

P is also defined as the electric dipole moment of material per unit volume.

P = np

where n is the number of molecules per unit volume.

Displacement vector, D is equal to the free charge per unit area or equal to the surface density of free charges,

Thus \(D = {q\over A} = σ \)    ---(3)

where q is free to charge and σ  is surface density of free charges.

As for the parallel plate capacitor:

\(E = {σ \over ε_0} \)   ---(4)

\(E_p = {σ_p \over ε_0}\)    ---(5)

By substituting equations 4 and 5 in equation 1, we get

\(E = {σ \over ε_0} – {σ_p \over ε_0}\) or ε0E = σ – σ0

By putting equations 2 and 3 in the above equation, we get

D = ε0E + P

Concept:

Electric field:

• An electric property associated with each point in space when charge is present in any form.
• The magnitude and direction of the electric field are expressed by the value of E, called intensity of electric field.
• The direction is away positive charge, and toward a negative one

​Intensity of electric field due to point charge at a distance r is given by 

 \(E = \frac{1}{4 \pi ϵ _∘} \frac {q}{r^2}\)

where ϵ_∘ = permittivity of free space

Explanation:

The electric field on axial line (at point P) is is given by

\(E = \frac{1}{4 \pi ϵ _∘} \frac {\underset{P}{\rightarrow}}{r^3}\)

Where

P = dipole moment vector

Hence from above equation we can see that, the electric field at x distance from the center of a dipole will be 1/x³.

Additional Information

Electric dipole: ​

• When two equal and opposite charges q are separated by a small distance d, then this combination of charges are called as electric dipole.
• The multiplication of charge and the distance between them is called as electric dipole moment.
• The electric dipole moment is denoted by P and the SI unit of dipole moment is Coulomb-meter (Cm)

Dipole moment

P = q × d

CONCEPT: 

• Electric Dipole moment: The measure of the polarity of the system of two electric charges is called an electric dipole moment.

• The simplest example of an electric dipole is a pair of electric charges of equal magnitude and opposite signs separated by distance.

At any point at r distance from the dipole, the Electric field

\(E = \frac{Kp}{r^3}\sqrt{3cos^2θ+1}\)

where p is the dipole moment, r is the distance from the dipole, θ is the angle, and K is the constant.

EXPLANATION:

At any point at r distance from the dipole, the Electric field

\(E = \frac{Kp}{r^3}\sqrt{3cos^2θ+1}\)

The electrostatic field at a point P on the axis of the dipole at a distance r from the midpoint of a dipole:

\(E = \frac{Kp}{r^3}\sqrt{3cos^2θ+1}\) and θ = 0° 

\(E = K\frac{2p}{r^3} = \frac{1}{4\pi \epsilon_0}\frac{2p}{r^3}\)

So the correct answer is option 2.

Concept:

Gauss law:

\( \oint_{\mathcal{S}} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \)

The surface integral is the total flux in a closed surface for Electric Field.

Using Gauss Law the following formula can be derived for the Electric field at distance r:

E = 2 kλ / r 

The direction of the electric field will be normal to the wire.

Calculation:

P) Electric Field is equal due to both wires in opposite directions.

Hence net electric field is 0.

Q) Electric Field is in the same direction due to both wires.

E_net = 2kλ /(d/2) + 2kλ /(d/2)

E_net = 8kλ /d

R) Electric Fields in the opposite direction

Enet = 2kλ /(d/2) - 2k(λ/2) /(d/2)

= 2kλ /d

S) Electric Field due to both wires in same direction :

 E_net = 2 kλ/(d/2) + 2k(λ/2)/(d/2)

Enet = 6kλ / d.

∴ Option 4) is correct

Concept:

Electric Field of an Electric Dipole

• Axial Line

The electric field at a point on the axial line (line extending through both charges of the dipole) is given by:

E_axial = \(\frac{1}{4\pi\epsilon_0}\frac{3p}{r^3}\)

Where:

p is the dipole moment (p = q × d, where q is the magnitude of each charge and d is the separation distance),
r is the distance from the center of the dipole to the point where the field is measured.

• Equatorial Line

The electric field at a point on the equatorial line (perpendicular bisector of the dipole) is given by:

E_equatorial = \(\frac{1}{4\pi\epsilon_0}\frac{p}{r^3}\)

Where:

p is the dipole moment,
r is the distance from the center of the dipole to the point where the field is measured.

Calculation:

⇒\(\mathrm{E}_{\mathrm{P}}=\frac{2 \mathrm{KP}}{\mathrm{r}^3}=\mathrm{E}\)

⇒\(E_R=\frac{K P}{(2 r)^3}=\frac{E}{16}\)

⇒x = 16 

The electric field at point R on equitorial line will be \(\rm \frac{E}{x}\) The value of x : 16