Notes-Part-1-Class-12-Physics-Chapter-8-Electrostatics-MSBSHSE

Electric Field Intensity due to Uniformly Charged Spherical Shell or Hollow Sphere:

Consider an isolated charged hollow spherical conductor A, of radius R and surface charge density σ, placed in a medium of permittivity ε.

Consider a point P outside the conductor at a distance r from its centre. To find the electric field intensity at P, we choose a spherical Gaussian surface S of radius r through P and concentric with conductor A. A small element of this surface containing P has an area dS.

The charge on the sphere is Q = σ (4πR²)  …..(1)

The charge Q is uniformly distributed over the outer surface of the spherical conductor. Then, by symmetry, the electric field intensity at every point on surface S is normal to the surface and has the same magnitude E. If charge Q is positive, \(\vec{E}\) at every point on S is radially outward.

The angle θ between \(\vec{E}\) and \(\vec{ds}\) being zero for every surface element, the electric flux throughevery element is

dΦ = \(\vec{E}.\vec{ds}\).  = E

Therefore, the flux through the Gaussian surface S is

Φ = \(\oint E.ds\) = (E \oint ds\)  ...(2)

\(\oint ds\) = surface area of the sphere = 4πR²

∴ Φ = E x 4πr²  (3)

Then, by Gauss’s theorem,

Φ = Q/ε = E x 4πr²  ...(4)

∴ E = \(\frac{Q}{4πεr^2}=\frac{Q}{4πε_0kr^2} \)   ……..(5)

where ε₀ is the permittivity of free space and k = ε/ε₀ is the relative permittivity (dielectric constant) of the surrounding medium.

∴ E = \(\frac{σ(4πεR^2)}{4πεr^2} \)= \(\frac{σR^2}{εr^2} \)   ……..(6)

Equations (5) and (6) give the magnitude of the electric field intensity at a point P outside a hollow spherical conductor. If the net charge Q enclosed by the Gaussian surface is positive,  is radially outward; if Q is negative, F is radially inward.

Equation (5) shows that for a point outside a hollow spherical conductor carrying a charge Q, the conductor behaves like a point charge Q at its centre.

Case (1) electric intensity on the surface of (i.e., just outside) the spherical conductor:

At a point just outside the sphere, r ≈ R

∴ E = σ/ε = σ/kε₀

Case (2) electric intensity inside the spherical conductor:

Since electric charge resides on the outer surface of a hollow conductor, the charge inside the hollow spherical conductor σ is zero. Then, E_inside = 0.

 Electric Field Intensity due to an Infinitely Long Straight Charged Wire:

Consider a uniformly charged wire of infinite length having a constant linear charge density λ (charge per unit length), kept in a medium of permittivity ε (ε = ε₀k ) .

To find the electric field intensity at P ,at a distance r from the charged wire, imagine a coaxial Gaussian cylinder of length l and radius r (closed at each end by plane caps normal to the axis) passing through the point P.

As λ is the charge per unit length of conductor A, the net charge enclosed by the Gaussian cylinder of length l is

Q = λl      ……..(1)

Consider a very small area ds at the point P on the Gaussian surface.

Charge is uniformly distributed over the outer surface of the cylindrical conductor. Then, by symmetry, the electric field intensity at any point outside the conductor is perpendicular to the cylinder axis. Hence, the component of the electric field intensity perpendicular to the plane circular faces of the Gaussian surface is zero. Therefore, the electric flux through these flat faces is zero.

By symmetry, the electric field intensity \(\vec{E}\) at every point on the curved face of surface S is normal to the surface and has the same magnitude E. If the charge on conductor A is positive, \(\vec{E}\) is directed along the outward drawn normal \(\vec{ds}\).

The angle θ between \(\vec{E}\) and \(\vec{ds}\) being zero for every surface element, the electric flux through every element is

dΦ = \(\vec{E}.\vec{ds}\)  = E ds

Therefore, the flux through the curved face of the Gaussian surface S is

Φ = \(\oint E.ds\)  = \(E\oint ds\)     ...(2)

\(E\oint ds\)= area of the curved surface = 2πrl, where l is the length of the cylinder as shown in the figure.

∴ Φ = E x 2πrl      ...(3)

Then, by Gauss’s theorem,

Φ = Q/ε  = E x 2πrl    ...(4)

∴ E = \(\frac{λl}{ε(2πrl)}\)=\(\frac{λ}{2πεr}\)= \(\frac{λ}{2πε_0kr}\)    ……….(5)

where ε₀ is the permittivity of free space and k = ε/ε₀ is the relative permittivity (dielectric constant) of the surrounding medium.

This gives the magnitude of the electric field intensity in terms of the linear charge density λ. For positive λ, \(\vec{E}\)  is outward, while for negative λ, \(\vec{E}\) is inward.

Remember :

(1) If σ is the surface charge density (charge per unit area),

σ = \(\frac{λl}{2πRl}\) = \(\frac{λ}{2πR}\)

∴ λ = 2πRσ

∴ E =\(\frac{2πRσ}{2πε_0kr}=\frac{Rσ}{ε_0kr} \)

(2) If the charge on the conductor is negative, \(\vec{E}\) is inward

Expression for force per unit area on a charged conductor  :

Consider a positively charged conductor of any shape placed in a dielectric medium of permittivity ε

Let dS be a very small area of the surface of the conductor drawn around point A as shown in the figure. If σ is the surface charge density on dS, the charge on the area element is σdS and the electric field intensity at a point just outside the area element has a magnitude E = σ/ε, \(\vec{E}\)  can be considered as a resultant of :

(i) \(\vec{E_1}\) due to the charge on the area dS and (ii) \(\vec{E_2}\) due to the charge on the remaining part of the conductor, i.e., excluding dS. Since the two fields have the same direction,

E = E₁ + E₂    ….(1)

\(\vec{E_1}\) has opposite directions on the two sides of dS; but on both sides, \(\vec{E_2}\) is directed along the outward normal to the element.

At point B inside the conductor, the electric field is zero.

∴ E₂ − E₁ = 0  E₂ = E₁

At A, just outside the conductor E₁ + E₂ = E

∴ E₂ = E₁ = ½ E

But, E = σ/ε     …..(2)

∴ E₂ = E₁ = σ/2ε    …..(3)

Thus, the charged area element σdS is situated in an electric field of magnitude E₂ = σ/2ε , produced by the charge on the rest of the conductor and this field is outward. The force acting on dS due to the action of this field.

= (charge on area dS) x (electric field due to the remaining charge)

∴ F = σdS x E₂ = \(σdS×\frac{σ}{2ε}=\frac{σ^2}{2ε}dS\)    …..(4)

This force acts outward, normal to the surface. The magnitude of the mechanical force per unit area of the surface is

f = F/ds = \(\frac{σ^2}{2ε}=\frac{σ^2}{2kε_0}\)    …..(5)

where ε₀ is the permittivity of free space and k = ε/ε₀ is the relative permittivity of the medium.

From Eq. (2), σ = ε E = ε₀kE

Substituting this value of σ in Eq. (5),

f =\frac{(ε_0kE)^2}{2kε_0}\) = \(\frac{1}{2}ε_0kE^2\)

Thus, the mechanical force per unit area of a charged conductor,

f = \(\frac{σ^2}{2kε_0}\) or f = \(\frac{1}{2}ε_0kE^2\)

If the charge on the conductor is negative, the field  at any area element will be inward. But charge on the area element being negative, the force    on the area element due to  will still be outward, normal to the surface,

Therefore, the force on the charged area element σdS is

F = ½ εE²ds. This force acts outward, normal to the surface.

Expression for work done in displacing a charged area element  :

Suppose an external agent displaces the charged area element through a small distance dx in a direction opposite to the above outward force, always keeping the element in equilibrium. To do so, the external agent must apply a force of equal magnitude in the opposite direction. Then, the work done by the external agent against the mechanical force,

dW = Fdx = (½ εE²ds) dx

During the displacement dx, the area element sweeps out a volume, dV= dS.dx.

∴ dW = ½ εE²dV

This work done is stored in the electric field in the medium in the form of potential energy. Therefore, the energy density, i.e., the energy stored per unit volume of the medium is

dW/dV = \(\frac{1}{2}εE^2\) = \(\frac{1}{2}ε_0kE^2\)

Expression for potential energy:

Consider a test charge q_o in the electric field \(\vec{E}\) of a source charge + Q. The electric force acting on the test charge, \(q_0\vec{E}\) , is a conservative force. When the test charge is moved in the field at constant velocity by some external agent, the work done by the field on the charge is equal to the negative of the work done by the external agent causing the displacement. Suppose an external agent moves the test charge without acceleration from a point A, at a distance r₁ from + Q, up to a point B, at a distance r₂, see Fig.

Since the electric field surrounding a point charge is not uniform, the electrostatic force on q_o increases as it approaches Q. Consequently, the external agent has to exert on q_o a force of increasing magnitude and, for equal displacements, do increasing amount of work. Because the force exerted varies along the path, we imagine the total displacement to be made up of a large number of infinitesimal displacements \(\vec{dx}\) . The distance dx is so small that, at an average distance x from Q, the electrostatic force \(\vec{F}\) on q_o has a constant magnitude

F = \(\frac{1}{4πε_0}\frac{Qq_0}{x^2}\)

over the distance dx. The force \(\vec{F_E}\) in by the external agent is equal and opposite to \(\vec{F}\)  at every instant : \(\vec{F_E}\) = \(-\vec{F}\)

Therefore, the infinitesimal work dW done by the external agent for the displacement \(\vec{dx}\) is

dW = \(\vec{F_E}.\vec{dx}\)  = F_E.dx = −Fdx

The total work done by the external agent in moving the test charge from B up to A is the line intergral of dW between the limits x = r₁ and x = r₂.

W = \(\int_{x=r_1}^{x=r_2}dw=\int_{r_1}^{r_2}(-\frac{1}{4πε_0}\frac{Qq_0}{x^2})dx\)

= \(-\frac{1}{4πε_0}Qq_0\int_{r_1}^{r_2}\frac{dx}{x^2}\)

= \(-\frac{1}{4πε_0}Qq_0[\frac{dx}{x^2}]_{r_1}^{r_2}\)

=\(-\frac{1}{4πε_0}Qq_0[-\frac{1}{r}-(-\frac{1}{r_1})]\)

= \(\frac{1}{4πε_0}Qq_0(\frac{1}{r}-(\frac{1}{r_1})\)  = ΔU

where ΔU = U_B — U_A is the change in the potential energy of the test charge in moving it from the point A to the point B. Choosing the potential energy of q_o to be zero when it is infinitely far away from Q, i.e., r₁ = ∞, its potential energy at a distance r from Q is

U(r) = \(\frac{1}{4πε_0}\frac{Qq_0}{r}\)

Thus, the potential energy U of the system of two point charges q₁ and q₂ separated by r can be obtained from the above equation by using r₁ = ∞, and r₂ = r,. It is then given by,

U(r) = \(\frac{1}{4πε_0}\frac{q_1q_2}{r}\)

Remember : Taking U(∞) =0 is often a convenient reference level in electrostatics, but in circuit analysis other reference levels are often more convenient.

Units of potential energy :

SI unit= joule (J)

“One joule is the energy stored in moving a charge of 1C through a potential difference of 1 volt.

Relation between electric field and electric potential:

Consider a test charge q_o as it moves from point A to point B in an electric field. The electric force on q_o at any point along the path is

\(\vec{F}\) = \(q_0\vec{E}\) ,  where \(\vec{E}\) is the field at that point.

The incremental work dW done by the field as q_o undergoes a displacement \(\vec{dl}\) along the path is

dW = \(\vec{F}.\vec{dl}\)  = \(q_0\vec{E}.\vec{dl}\)= q_o E dl

In the process, the charge q_o is moved from a higher potential to a lower potential, thereby losing potential energy. Therefore, the change in potential energy

dU = −dW = −q_o E dl

By definition, the change in electric potential is

dV = dU/q₀

∴ dV = − E dl

∴ E = − dV/dl

If E_x, E_y and E_Z are the rectangular components

E_x = − dV/dx ,  E_y = − dV/dy,   E_z = − dV/dz

The quantity dV/dl  is the rate at which the electric potential changes with distance and is called the electric potential gradient. The above equation thus shows that the magnitude of the electric field intensity at a point is equal to the negative of the potential gradient at that point.

Electric potential due to an electric dipole:

Consider an electric dipole AB of dipole length 2l and point charges + q and −q. Its electric dipole moment \(\vec{p}\) has magnitude p = 2ql.

Let C be a point at a distance r from O, the centre of the dipole, in a direction θ with the dipole axis, as shown in Fig.

Let AC = r₁ and BC = r₂.

The electric potential at P due to the charge + q is

V₁ = \(+\frac{1}{4πε_0}\frac{q}{r_1}\)    …..(1)

and that due to the charge −q is

V₂ = \(-\frac{1}{4πε_0}\frac{q}{r_2}\)    …..(2)

Since potential is a scalar quantity, the resultant electric potential at P due to the dipole is

V= V₁ + V₂ = \(+\frac{1}{4πε_0}\frac{q}{r_1}\) \(-\frac{1}{4πε_0}\frac{q}{r_2}\)  

= \(\frac{q}{4πε_0}(\frac{1}{r_1}-\frac{1}{r_2})\) = \(\frac{q}{4πε_0}(\frac{r_2-r_1}{r_1r_2})\)       …..(3)

In the figure, CA₁ = CA = r₁ and CB₁ = CB = r₂.

For short dipole (2l << r), ∠OA₁A ≅ ∠OB₁B ≅ 90°.

Hence, in the right angled Δ  OA₁A and OB₁B,

OA₁ = OB₁ = lcos θ

r₂ — r₁ = CA₁ − CB₁ = A₁B₁ = 2lcos θ    ……(4)

Also, r₁ r₂ = CA₁ x CB₁ = (CO + OA₁)(CO − OB₁)

= (r + lcos θ)(r − lcos θ)

= l² – l²cos²θ

∴   r₁ r₂ ≈ r²      (‘.’ l << r)   …..(5)

From eq. (3), (4) & (5)

V = \(\frac{q}{4πε_0}(\frac{2lcosθ}{r^2})\)=\(\frac{1}{4πε_0}(\frac{(2ql)cosθ}{r^2})\)

= \(\frac{1}{4πε_0}(\frac{pcosθ}{r^2})\)     …..(5)

This is the required expression

Particular cases :

(i) At a point on the dipole axis, θ = 0° (nearer to the charge +q) or 180° (nearer to the charge −q)

Cos θ = ± 1

∴ V_axis= ±\(\frac{1}{4πε_0}(\frac{pcosθ}{r^2})\)

(ii) At a point on the dipole equator, θ = 90° or 270°.

cos θ = 0 ∴ V_equator = 0

Remember : Since V_equator = 0, the equatorial plane of a dipole is an equipotential plane of electric potential equal to zero. No work is required to move a charge anywhere in the equatorial plane

Electrostatics potential due to a system of charges :

Or

Expression for the electric potential at a point due to several point charges :

Consider a system of charges q₁, q₂ ......... q_n at distances r₁, r₂ ...... r_n respectively from point P. The potential V₁ at P due to the charge q₁ is

V₁ = \(\frac{1}{4πε_0}\frac{q_1}{r_1}\) 

Similarly the potentials V₂, V₃ ........V_n at P due to the individual charges q₂, q₃ ...........q_n are given by

V₂ = \(\frac{1}{4πε_0}\frac{q_2}{r_2}\), V₃ = \(\frac{1}{4πε_0}\frac{q_3}{r_3}\) , …….. V_n = \(\frac{1}{4πε_0}\frac{q_n}{r_n}\) 

By the superposition principle, the potential V at P due to the system of charges is the algebraic sum of the potentials due to the individual charges.

∴ V = V₁ + V₂ +......+ V_n

= \(\frac{1}{4πε_0}(\frac{q_1}{r_1}+\frac{q_2}{r_2}+....\frac{q_n}{r_n})\)

=\(\frac{1}{4πε_0}\sum_{i=1}^{n}\frac{q_i}{r_i}\)

For a continuous charge distribution, summation should be replaced by integration.

Remember : Electric potential is a scalar quantity. To calculate the resultant potential due to two or more point charges, the potentials due to individual charges are added as simple scalars along with its sign, determined by the sign of the q that produces V