Field Theory - Electric Dipole....

- An electric dipole consists of two point charges of equal magnitude but of opposite sign and separated by a small distance.

- Consider an electric dipole centered at origin and placed in z – axis as shown in the figure:

- The potential (V) at point P is given as:

- If the distance between the charges (d) is very small as compared to the distance of the point P from the origin i.e.

If r >> d,

r₂ – r₁ ≅ d cosθ ;          r₁ ≅ r₂ = r ;           r₁r₂ ≅ r²

Substituting the values in the above equation, the potential at point P becomes:

- Electric field intensity (E) is the negative gradient of Electric Potential (V).
Hence,

- The expressions for electric potential (V) and field intensity (E) above are only valid for a dipole centered at the origin and aligned with the z-axis.

- To determine the fields produced by any arbitrary location and alignment, we first need to define a new quantity p, called the Dipole Moment.

p = Q d

- Since the distance d is a vector quantity, the dipole moment p is also a vector quantity.

- Dipole moment p is a measure of the strength of the dipole and indicates its direction.

- Vector d is a directed distance that extends from negative charge (- Q) to positive charge (+ Q). This directed distance vector d thus describes the distance between the dipole charges, as well as the orientation of the charges.

Therefore

d = | d | a_d

Where | d | is the distance between the charges and a_d defines the orientation or direction of the dipole.

- Say a dipole is aligned along z – axis, then directed distance d is given as:

d = | d | a_z

From the above diagram it’s clear that:

a_z . a_r = cosθ

Hence the expression can be written as:

Q d cosθ  =  Q | d | a_z . a_r  =  Q d. a_r  =  p . a_r




- Hence electric potential (V) due to a electric dipole centered at origin and aligned with the z axis is rewritten as:

- The above expression no doubt is applicable for all and any dipole moments p, but is valid for dipoles centered at origin.

- Electric potential (V) at point P with a position vector r due to a dipole centered at a point with position vector r₁ is given as:

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- Energy Density In Electrostatic Field / Work Done To Assemble Charges.....


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Field Theory - Electric Potential (V) Due To A Circular Disc...

- Consider a circular disc of radius ‘a’ which carries a uniform surface charge density ρ_s , C /m².

- Say the disk lies on x - y plane (or z = 0 plane) with its axis along the z axis as shown in the figure.

- We need to find out electric potential (V) due to a circular disk at a point P (0, 0, h) on the z axis (z > 0).

 - Electric potential (V) at a point due to any surface charge (ρ_s) is given as:

- In this case,

ds = ρ dρ dφ

(Since it’s a disc, the varying terms are radius ρ and angle φ)

R = (ρ² + h²)^1/2

- Hence electric potential (V) is given as:

- On solving further the equation becomes

- As a → 0, electric potential (V) also tends to zero i.e. V → 0.

- Hence the electric potential at point (0, 0, h) is given as:

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- Electric Dipole....

- Numericals/Solved Examples - Electric Potential, Potential Gradient and Electric dipole...


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Field Theory - Relationship Between Electric Field Intensity(E) and Electric Potential(V)...

- The work done per unit charge in moving a test charge from point A to point B is the electrostatic potential difference between the two points(V_AB).

V_AB = V_B - V_A

Similarly,

V_BA = V_A – V_B

- Hence it’s clear that potential difference is independent of the path taken.
Therefore

V_AB = - V_BA

V_AB + V_BA = 0

- ∫_A^B (E . dl) + [ - ∫_B^A (E . dl) ] = 0

- The above equation shows that the line integral of Electric field intensity (E) along a closed path is equal to zero.



- In simple words,

“No work is done in moving a charge along a closed path in an electrostatic field”.

- Applying Stokes’ Theorem to the above Equation, we have:

- If the Curl of any vector field is equal to zero, then such a vector field is called an Irrotational or Conservative Field.

- Hence an electrostatic field is also called a conservative field.

- The above equation is called the second Maxwell’s Equation of Electrostatics.

- Since Electric potential is a scalar quantity, hence dV (as a function of x, y and z variables) can be written as:

- Hence the Electric field intensity (E) is the negative gradient of Electric potential (V).

- The negative sign shows that E is directed from higher to lower values of V i.e. E is opposite to the direction in which V increases.




EQUIPOTENTIAL SURFACE:

- An equipotential surface refers to a surface where the potential is constant.

- The intersection of an equipotential surface and a plane results into a path called an equipotential line.

- No work is done in moving a charge from one point to the other along an equipotential line or surface i.e. V_A – V_B = 0

Hence,

From the above equation, it’s clear that the electric flux lines and the equipotential surface and normal to each other.

- Because the electric field is the negative gradient of electric potential, the electric field lines are everywhere normal to the equipotential surface and points in the direction of decreasing potential.



- The equipotential lines for a positive point charge. The solid lines show the flux lines or electric lines of force.

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- Electric Dipole....

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Field Theory - Scalar Electric Potential / Electrostatic Potential (V)...

- If a charge is placed in the vicinity of another charge (or in the field of another charge), it experiences a force.

- If a field being acted on by a force is moved from one point to another, then work is either said to be done on the system or by the system.

- Say a point charge Q is moved from point A to point B in an electric field E, then the work done in moving the point charge is given as:

 W_A→B = - ∫_A^B (F . dl) = - Q ∫_A^B(E . dl)

where the – ve sign indicates that the work is done on the system by an external agent.

- The work done per unit charge in moving a test charge from point A to point B is the electrostatic potential difference between the two points(V_AB).

V_AB = W_A→B / Q 

= - ∫_A^B(E . dl) 

= - ∫_Initial^Final (E . dl)

- If the potential difference is positive, there is a gain in potential energy in the movement, external agent performs the work against the field.

- If the sign of the potential difference is negative, work is done by the field.

- The electrostatic field is conservative i.e. the value of the line integral depends only on end points and is independent of the path taken.

- Since the electrostatic field is conservative, the electric potential can also be written as:



V_AB = - ∫_A^B (E . dl ) 

= - ∫_A^P_o (E . dl) - ∫ _Po^B (E . dl)

= - ∫_Po^B (E . dl)  -  (- ∫_A^P_o(E . dl)

= V_B – V_A

Thus the potential difference between two points in an electrostatic field is a scalar field that is defined at every point in space and is independent of the path taken.

- The work done in moving a point charge from point A to point B can be written as:

W_A→B = - Q [V_B – V_A] = - Q ∫_A^B (E . dl)

- Consider a point charge Q at origin O.

- Now if a unit test charge is moved from point A to Point B, then the potential difference between them is given as:

- Electrostatic potential or Scalar Electric potential (V) at any point P is given by:

V = - ∫_Po^P (E . dl)

The reference point P_o is where the potential is zero (analogues to ground in a circuit).

- The reference is often taken to be at infinity so that the potential of a point in space is defined as

V = - ∫_∞^P (E . dl)

- Basically potential is considered to be zero at infinity. Thus potential at any point ( r_B = r) due to a point charge Q can be written as the amount of work done in bringing a unit positive charge from infinity to that point (i.e. r_A → ∞)

- Electric potential (V) at point r due to a point charge Q located at a point with position vector r₁ is given as:

- Similarly for N point charges Q₁, Q₂ ….Q_n located at points with position vectors r₁, r₂, r₃…..r_n, the electric potential (V)  at point r is given as:

- The charge element dQ and the total charge due to different charge distribution is given as:

dQ = ρ_ldl    → Q = ∫_L (ρ_ldl) → (Line Charge)

dQ = ρ_sds   → Q = ∫_S (ρ_sds) → (Surface Charge)

dQ = ρ_vdv    → Q = ∫_V (ρ_vdv) → (Volume Charge)

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- Electric Potential (V) Due To A Circular Disc...


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Field Theory - Numericals/Solved Examples - Gauss's law...

Q.1 A point charge of 30 nC is located at the origin while plane y = 3 carries charge 10 nC / m² . Find D at (0, 4, 3)?
Ans:

Electric flux density (D) at point (0, 4, 3) due to a point charge and line charge is given as:

D = D_Q + D_ρ

= 0.019 (0, 4, 3) + 5a_y

= 0.76 a_y + 0.057 a_z + 5a_y

= 5.076 a_y + 0.057 a_z




Q.2 A charge distribution in free space has ρ_v = 2r nC / m³ for 0 < r < 10 m and zero otherwise. Determine E at r = 2m and r = 12m?
Ans:

Gauss’s law states that:

For 0 ≤ r ≤ 10

D_r (4πr²) = ∫ ∫ ∫ 2r (r² sinθ dr dθ dφ)

D_r (4πr²) = 4π (2r⁴) / 4

For r ≥ 10

D_r (4πr²) = ∫ ∫ ∫ 2r_o (r² sinθ dr dθ dφ)

D_r (4πr²) = 4π (2r_o⁴) / 4

D_r (4πr²) = 2πr_o⁴

Q.3 If D = (2y² + z)a_x + 4xy a_y + x a_z C/m², find
a) Volume charge density at (-1, 0, 3)
b) The flux through the cube defined by 0 ≤ x ≤ 1,
c) The total charge enclosed by the cube.
 Ans:

a) Volume charge density (ρ_v) is defined as:

ρ_v = ∇ . D

   = 4

At point (-1, 0,3) ρ_v = 4

b) The total charge (Q) enclosed by the cube

Q = ∫_v ρ_v dv

= ∫_(x=0)¹ ∫_(y=0)¹ ∫_(z=0)¹ 4x (dxdydz)

= 4 | (x² / 2) |₀¹ (1) (1)

= 2 C

c) The flux through the cube defined by:

Ψ = Q_enc = 2C




Q.4 If the volume charge density (ρ_v) of a given charge distribution is given by ρ = ρ_o (a / r) in spherical co-ordinate, determine the electric flux density (D) at any point ?

Ans:

Gauss’s law states that:

Q_enc = ∫_v ρ_v dv

= ∫(r=0)^r ∫_(θ=0)^π ∫_(φ=0)^2π  ρ_o (a/r) (r² sinθ dr dθ dφ)

= - ρ_o a | r² / 2 |₀^r (2π) | (cosθ) |₀^π

= - ρ_o a (r² / 2) (2π) ( -2)

= 2 ρ_o aπr²

Electric flux is given as:

Since,

Ψ = Q_enc

D_r (4πr²) = 2 ρ_o aπr²

D = [ (ρ_o a) / 2a ] a_r


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Field Theory - Gauss Law - Application To A Uniformly Charged Sphere...

- Consider a sphere of radius “a” having a uniform volume charge density of ρ_o C/m³.

- Gaussian surface selected for a symmetric sphere charge is a sphere itself.

- Here we consider two cases:
a) A Gaussian surface with a radius r < a.

b) A Gaussian surface with a radius r > a.

CASE 1:

If r < a (consider the 1st figure), the total charge enclosed by the Gaussian surface is:

Electric flux is given as:

Here ds is taken to be a function of θ and φ( since the surface is a hollow sphere).

Gauss law states that Ψ = Q_enc

Therefore,

D_r 4π r² = (4 / 3) (ρ_o πr³)

              Or

D = (r / 3) ρ_o a_r (0 < r < a) 



CASE II: 

If r > a (consider the above figure), the total charge enclosed by the Gaussian surface is:

Electric flux is given as:

 Gauss law states that Ψ = Q_enc



Therefore,
D_r 4π r² = (4 / 3) (ρ_o πa³)

              Or

D = (a³ / 3r² ) ρ_o a_r (r > a)


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Field Theory - Gauss Law - Application To An Infinite Line & Sheet Charge...

- Consider an infinite line charge carrying a charge per unit length of ρ_L along the z axis.

- Gaussian surface selected for a symmetric line charge is a hollow cylinder of radius ρ and length l as shown in the figure

- A cylinder has basically three surfaces: top, bottom and the curved cylindrical surface.

From the diagram its clear that

- Electric flux density (D)  is parallel to the top and bottom Gaussian surface.

- Electric flux density (D) is normal to the curved cylindrical Gaussian surface.




- Differential surface (ds) is always normal to a surface.

D and ds are normal to each other for the top and bottom Gaussian surface.

Hence

∫ (D . ds) = 0

Hence it’s clear that D and ds are parallel to each other only for the curvilinear Gaussian surface.

- Since the Gaussian surface is a hollow cylinder, hence the variable terms are φ and z. Thus the differential surface for a hollow cylinder is given as:

ds = ∫_(φ=0)^2π ∫₀^l  (ρ dφ dz) a_ρ → S = 2πρl a_ρ

Hence applying gauss law, we have

→ Q_enc = D_ρ 2πρl = ρ_L l

INFINITE SHEET OF CHARGE

- Consider an infinite sheet charge carrying a charge per unit surface of ρ_s on a x-y or z = 0 plane.

- Gaussian surface selected for a symmetric sheet charge can be either a cylinder box or a rectangular box.

- The Gaussian surface is placed such that two of its faces are parallel to the sheet.

- In this case say the Gaussian surface is a rectangular box.

- From the diagram it’s very clear that only two faces of the rectangular box is parallel to the sheet and also parallel to the z-axis.

- Applying Gauss law, we have

 











- The electric field(E) as well as electric flux density(D) both points away from the plane if ρs is positive and towards the plane if ρs is negative.

- The magnitude of the electric field is a constant – the magnitude is independent of the distance from the infinite plane.



- This is because no matter how far the point is from the infinite sheet, the distance becomes incomparable with the dimensions of the plane. Hence it seems the point is very close to the infinite plane.


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- Numericals / Solved Examples - Gauss's law...


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Field Theory - Gauss Law (Theory) & Application To A Point Charge...

- Gauss law is one of the fundamental laws of Electrostatics.

- It states that

“The net electric flux emanating or coming from a close surface S is equal to the total charge contained within the volume V bounded by that surface.”

We know,

Therefore

Ψ = Q_enc

Hence Gauss law also states that

 “The total electric flux Ψ through any closed surface is equal to the enclosed electric charge.”

- Charge contained in a volume is given as:
dQ = ρ_v dv → Q = ∫_V ρ_v dv

Hence we have,

 


Applying Divergence theorem we have,

Comparing the above two equations, we have

And hence 

∇ . D = ρ_v

This equation is called the 1^st Maxwell's equation of electrostatics.

- Gauss law is an easy way of finding electric field for some symmetric problems in electrostatics.

- Gauss law relates the electric field at points on a closed Gaussian surface to the net charge enclosed by that surface.

- One thing to remember is Q_enc contains charges which are enclosed within the volume. Charges outside the volume, no matter how large or how close it may be, are not included in the term Q_enc.




Procedure To Apply Gauss Law:

- Check out whether symmetric charge distribution exists or not.

- If yes, a hypothetical surface called a Gaussian surface is selected such that E / D (due to charge) is either normal or tangential to the Gaussian surface.

- Gaussian surface is a hypothetical (any imaginary) closed surface enclosing the charge configuration.

- Gaussian surface is a surface to which the electric flux density is normal and over which equal to a constant value.




Application Of Gauss Law To A Point Charge:

- Consider a point charge Q at the origin, say at point P, the electric flux density (D) is to be evaluated.

- Gaussian Surface selected for a point charge is a sphere centered at origin.

From the diagram its clear that

- Electric flux density is everywhere normal to the Gaussian surface i.e. D = D_r a_r

- The total charge enclosed by the Gaussian surface, Q_enc = Q.


- Since the Gaussian surface is a hollow sphere, hence the variable terms are θ and φ. Thus the differential surface for a hollow sphere is given as:

ds = ∫_(φ=0)^2π ∫_(θ=0)^π  (r² sinθ dθ dφ) a_r = 4πr² a_r

- Hence applying gauss law, we have

→ Q_enc = D_r 4πr²

Hence

Again we know that, D = ε E.

Therefore

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- Gauss Law - Application To An Infinite Line & Sheet Charge...

- Gauss Law - Application To A Uniformly Charged Sphere...

- Numericals / Solved Examples - Gauss's law...



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