Electrostatics
Applications of coulombs law, Gauss’s Law and applications
Electric Field due to a Dipole
Two equal and opposite charges separated by a small distance form an electrical dipole.
Consider two point charges +q and –q of equal magnitude lying distance d apart.
The electric field intensity E due to a dipole at point P. The point P is at a distance x along the perpendicular
bisector of the line joining the charges.
Let the electric field intensities at point P due to the charges +q and –q be E+ and E- respectively.
The total electric field intensity at point P due to the charges +q and –q is.
E=E++EFor the present case, |E+|=|E-| because the point P is equidistant for the charges +q and –q . Therefore,
1
|E+|=|E-|=4𝜋𝜀
𝑞
2
0𝑟
1
𝑞
= 4𝜋𝜀
2 𝑑ൗ 2]
2
0 [𝑥 +
-------------------------1
x-components of E+ and E- will cancel the effect of each other, while the y- components of E+ and E- added up
to give of resultant electric field intensity E of electric dipole. Therefore,
|E|=|E+|cosθ +|E+|cosθ = 2|E+|cosθ =
From figure cosθ =
𝑑ൗ
2
𝑟
𝑑ൗ
2
⇒ 𝑐𝑜𝑠𝜃 =
𝑥 2 +(𝑑ൗ2)2
• By putting the value of cosθ in eq. 2 we get
𝑑ൗ
1
𝑞 𝑐𝑜𝑠𝜃
𝑑
2
• |E|=2|E+|
=
2 𝑑ൗ 2
𝑥 2 +(𝑑ൗ2)2
4𝜋𝜀0 [𝑥 +
2
]
𝑥 2 +(𝑑ൗ2)2
2
𝑞 𝑐𝑜𝑠𝜃
--------------2
4𝜋𝜀0 [𝑥 2 + 𝑑ൗ2 2]
|E|=
1
𝑞
4𝜋𝜀0 [𝑥 2 + 𝑑ൗ2 2]
𝑑
=
𝑥 2 +(𝑑ൗ2)2
1
𝑞𝑑
4𝜋𝜀0 𝑥 2 + 𝑑ൗ2
/
2 3 2
The quantity p=qd is called dipole moment.
Therefore,
1
𝑝
/
|E|=
2 3 2
2
𝑑
4𝜋𝜀0 𝑥 + ൗ2
1
𝑝
|E|=4𝜋𝜀 𝑥3 1 𝑑 2 3/2
1+ ൗ𝑥2 ൗ2
0
− 3/2
𝑝
𝑑
2
|E|=
1 + Τ2𝑥
3
4𝜋𝜀0 𝑥
• By using binomial expansion, we get
𝑝
−3Τ ) 𝑑Τ
2+. . .
|E|=
1
+
(
2
2𝑥
3
4𝜋𝜀0 𝑥
• Neglect the 2nd and higher terms, then
𝑝
E=4𝜋𝜀 𝑥3
0
This is the expression of electric field due to a dipole.
Electric Field Intensity due to an Infinite Line of
Charges
• Consider a portion of an infinite line of positive charge
• The electric field intensity at point P at a perpendicular distance x from the
line of charge.
• As the charge is distributed uniformly over it, so it has constant linear
charge density μ;
𝐶ℎ𝑎𝑟𝑔𝑒 𝑞
μ=
=
𝑙𝑒𝑛𝑔𝑡ℎ
𝐿
• For an infinitesimal length element dy having charge dq.
𝑑𝑞
μ = ⇒ dq=μdy
𝑑𝑦
• The electric field intensity due to this length element at point P is given by;
1 𝑑𝑞
1 𝜇𝑑𝑦
dE=
=
4𝜋𝜀0 𝑟 2 4𝜋𝜀0 𝑟 2
1
𝑑𝑞
𝑑𝐸 =
4𝜋𝜀0 𝑥 2 + 𝑦 2
• The rectangular components of dE are dEx=dE cosθ and dEy=dE sinθ
• If we consider identical charge elements symmetrically located on both sides of O, then the ycomponents of electrical field intensity will cancel out each other and its x-components are added up to
give the total electric field intensity due to this continuous charge distribution.
Therefore, the total electric field intensity will be
𝑦=∞
𝑦=∞
𝑦=∞
𝐸 = ‫=𝑦׬‬−∞ 𝑑𝐸𝑥 = ‫=𝑦׬‬−∞ 𝑑𝐸 𝑐𝑜𝑠𝜃 = 2 ‫=𝑦׬‬0 𝑑𝐸 cos θ
∞
1 𝑑𝑞 𝑐𝑜𝑠𝜃
𝑑𝐸 = 2 න
2
2
0 4𝜋𝜀0 𝑥 + 𝑦
∞
−∞
1 𝜇 𝑑𝑦 𝑐𝑜𝑠𝜃
1
𝜇 𝑑𝑦 𝑐𝑜𝑠𝜃
𝑑𝐸 = 2 න
2 + 𝑦 2 ⇒⇒⇒ 𝐸 = 2𝜋𝜀 න
4𝜋𝜀
𝑥
𝑥2 + 𝑦2
0
0 0
0
𝜋/2
𝜇
𝑥 𝑠𝑒𝑐2𝜃 𝑑𝜃 𝑐𝑜𝑠𝜃
𝐸=
න
2𝜋𝜀0 0
𝑥 2 + 𝑥2 𝑡𝑎𝑛2𝜃
From fig. tanθ=y/x y= x tanθ
dy=x sec2 θ dθ
When x=0, θ=0
When x=∞, θ=π/2
𝜋/2
𝜇
𝜇
/
𝐸=
න 𝑐𝑜𝑠𝜃 𝑑𝜃 =
𝑠𝑖𝑛𝜃 |0 𝜋 2
2𝜋𝜀0 𝑥 0
2𝜋𝜀0 𝑥
𝜇
𝐸=
2𝜋𝜀0 𝑥
This is the expression for electric field intensity at point P due to an infinite line of charge.
Electric Field Intensity due to a Ring of Charge
• Consider a ring of positive charge of radius R. The electric field intensity at point P which is at the
distance r from the plane of ring. As the charge is distributed uniformly over it, so it has constant
linear charge density . For an infinitesimal length element ‘ds’ of ring, linear charge density μ;
𝐶ℎ𝑎𝑟𝑔𝑒 𝑑𝑞
𝑑𝑞
μ= 𝑙𝑒𝑛𝑔𝑡ℎ = 𝑑𝑠
μ = 𝑑𝑠 ⇒ dq=μds
• The electric field intensity due to this length element at point P is given by;
1 𝑑𝑞
1 𝜇𝑑𝑠
dE =
=
4𝜋𝜀0 𝑟 2 4𝜋𝜀0 𝑟 2
1
𝜇𝑑𝑠
𝑑𝐸 =
4𝜋𝜀0 𝑧 2 + 𝑅 2
• The rectangular components of dE are dEz=dE cosθ and dEy=dE sinθ
• If we consider the identical charge elements located on the opposite end of the diameter, then
dEy-components will cancel out each other and dEz-components are added up to give the final
value of electric field intensity at point P. Therefore, the total electric field intensity will be
𝐸 = න𝑑𝐸𝑧 = න𝑑𝐸𝑐𝑜𝑠𝜃
1 𝜇𝑑𝑠 𝑐𝑜𝑠𝜃
𝐸=න
4𝜋𝜀0
𝑟2
1
𝜇 𝑑𝑠 𝑧
𝐸=
න 2
4𝜋𝜀0
𝑟 𝑟
From fig. r2=z2+R2
r3=(z2+R2)3/2
𝑡𝑜𝑡𝑎𝑙 𝐶ℎ𝑎𝑟𝑔𝑒 𝑑𝑞
μ=
=
𝑡𝑜𝑡𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
2𝜋𝑟
μ(2𝜋𝑟) = q
𝜇𝑍
𝑑𝑠
𝜇𝑧
𝜇𝑧(2𝜋𝑟) 𝑧(𝜇 2𝜋𝑟)
𝐸=
=
න 3 ⇒𝐸=
න𝑑𝑠 =
3
3
4𝜋𝜀0 𝑟
4𝜋𝜀0 𝑟
4𝜋𝜀0 𝑟
4𝜋𝜀0 𝑟3
𝑧𝑞
E=
/2
2
2
3
4𝜋𝜀0 (𝑧 +𝑅 )
This is the expression for electric field intensity at point P due to charge ring.
• Special Case:
When the point P is far away from the ring, i.e., z >>R so that R2 can be neglected
𝑧𝑞
𝑞
E=
/2 =
2
3
4𝜋𝜀0 (𝑧 )
4𝜋𝜀0 𝑧2
• Which is the expression for the electric field intensity, when the field point is far away from the
ring. Thus, the charged ring acts like a point charge when the field point is at the large distance.
Electric Field Intensity due to a Disk of Charge
Consider a circular disk of uniform surface charge density ‘σ’.
The electric field intensity at point P which is at the distance z from the
plane of disk.
Consider a small element of the disk in the ring shape of radius w and
width dw. If dq is the charge on this element of ring having surface area
A, then,
𝐶ℎ𝑎𝑟𝑔𝑒 𝑑𝑞
σ=
=
𝐴𝑟𝑒𝑎
𝑑𝐴
dq=σ dA
dA=2πω dω
• The electric field intensity due to ring of charge at point P is given by;
𝑧𝑞
E=
/2
2
2
3
4𝜋𝜀0 (𝑧 +𝜔 )
𝑧𝑑𝑞
𝑧 σ 2πω dω
dE =
/2 =
/2
2
2
3
2
2
3
4𝜋𝜀0 (𝑧 +𝜔 )
4𝜋𝜀0 (𝑧 +𝜔 )
𝑧𝜎 2ω dω
dE =
/
4𝜀0 (𝑧 2 +𝜔2 )3 2
𝑑𝐸 =
3
−
𝑧𝜎(𝑧 2 +𝜔2 ) 2 2ω
𝜔=𝑅
4𝜀0
dω
3
𝑧𝜎
−
𝐸=
න
(𝑧 2 +𝜔2 ) 2 2ω dω
4𝜀0 𝜔=0
1
𝐸=
−𝑧𝜎
1
1
[ 2 2 − 2]
2𝜀0
𝑧
(𝑧 +𝑅 )
−2𝑧𝜎 2
−
𝐸=
(𝑧 +𝜔2 ) 2 |0R
4𝜀0
𝜎
𝑧
𝐸=
1−
2𝜀0
(𝑧 2 +𝑅2 )
This is the expression for electric field intensity at point P due to disc of charge.
• Special Case:
𝑧
0
if R>>z i.e. z=0 then 2 2 = 2 2 =0
𝑧 +𝑅
0 +𝑅
𝜎
𝐸=
2𝜀0
It means when , the disk of charge behaves like infinite sheet of charge.
Torque on a Dipole in a Uniform Electric Field
• Consider an electric dipole consists of +q and –q separated by a distance d.
• When an electric dipole is placed in an external electric field, the force on
the positive charge will be in one direction and the force on the negative
charge in another direction.
• The force on +q and –q have the equal magnitude but opposite in direction.
Therefore the net force on the dipole due to external field is zero. The
magnitude of each force is
• |F1|= |F2|=qE
• These two forces make a couple and so the torque acts on the dipole. The
magnitude of this torque is given as,
Torque = ( Force )x ( Moment Arm )
τ=F(dsinθ )=qE dsinθ
• τ = qd E sinθ
• τ = pE sinθ
‫ ؞‬dipole moment p=qd
In vector form, τ = p x E
• The direction of torque of dipole in a uniform electric field is determined by
right hand rule.
Energy of Dipole in a Uniform Electric Field
• Consider an electrical dipole is placed in a uniform electric field. We want to calculate the work done by the
electric field in turning the dipole through an angle θi. The work done by the electric field in turning the dipole
from an initial angle θi to the final angle θf is given by;
W = න 𝑑𝑤
𝜃𝑓
W = − න 𝜏 𝑑𝜃
𝜃𝑖
Where τ is the torque exerted by the electric field. The negative sign is necessary because the torque tends to
decrease θ. Also
τ = pE sinθ
𝜃𝑓
W = − න 𝑝𝐸 𝑠𝑖𝑛𝜃 𝑑𝜃
𝜃𝑖
• Assuming that the dipole is revolve from initial
angle θi=π/2 to the𝜃final angle θf=θ . Then
𝜃
W = − න𝜋 𝑝𝐸 𝑠𝑖𝑛𝜃 𝑑𝜃 = −pE න 𝑠𝑖𝑛𝜃 𝑑𝜃
2
𝜋/2
W=pE [cosθ –cos π/2 ]
W=pE cosθ
• Since the work done by the agent that produces the external field is equal to the negative of the change in
potential energy U of the system.
• U=-W= -pE cosθ
• Therefore, Potential Energy U=- p.E
• This is the expression of potential energy of the dipole in a uniform electric field
• Electric Flux: The number of electric lines of force passing
normally through a certain area is called the electric flux.
• It is measured by the product of area and the component of
electric field intensity normal to the area. It is denoted by
the symbol φe.
• Consider a surface S placed in a uniform electric field of
intensity E. Let A be the area of the surface. The component
of E normal to the area A.
Then, the electric flux through the surface is given by
φe= A (E cosθ)
φe=E A cosθ
φe=E. A
• Thus the electric flux is the scalar product of electric field
intensity and the vector area.
• The SI unit of the electric flux is Nm2/C
Electric flux through an irregular shaped object
• Consider an object of irregular shape placed in a non-uniform electric field.
• Divide the surface into n number of small patches having area ∆A1, ∆A2, ∆A3
….∆An.
• Let E1, E2, E3, ………En are the electric field intensities which makes angle θ1, θ2,
θ3, ……….θn with the normal to the area elements ∆A1, ∆A2, ∆A3 ….∆An
respectively.
• If be the electric flux φ1, φ2, φ3 …….φn, through ∆A1, ∆A2, ∆A3 ….∆An then the
total electric flux φe will be,
φe = φ1 + φ2+ φ3 …….+φn
• φe=E1(∆A1 cosθ1)+E2(∆A2 cosθ2)+E3(∆A3 cosθ3)+…….En(∆An cosθn)
φe=E1.∆A1+E2.∆A2+E3.∆A3+…….En.∆An
• Where, ∆A1, ∆A2, ∆A3 ….∆An are the vector areas corresponding to the area
elements ∆A1, ∆A2, ∆A3 ….∆An respectively.
φe =σ𝑛𝑖=1 Ei ∙ ∆Ai
• When n --> ∞ then the sigma is replaced by surface integral i.e.
φe =‫ 𝑠׬‬E ∙ dA
• By convention the outward flux is taken as positive and inward flus is taken
negative.
Gauss’s Law
• Differential form of gauss’s law
𝑑𝑞
𝑣𝑜𝑙𝑢𝑚𝑒 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦𝜌 =
𝑑𝑣
The total electric flux through any close surface is 1/ε0
times the total charge enclosed by the surface.
The Gauss’s law gives the relation between total flux
and total charge enclosed by the surface.
Consider a collection of positive and negative charges
in a certain region of space. According to Gauss’s law:
𝑞
φe =𝜀 --------------1
0
Where, q is the net charge enclosed by the surface.
Also,
ϕ𝑒 = ර E. dA −−− −2
• Comparing (1) and (2), we have:
𝑞
ර E. dA =
𝜀0
• The surface normal integral of electric field intensity
is equal to 1/ε0 times the total charge enclosed by
the surface.
𝑞 = න 𝜌 𝑑𝑣
𝑣
By Gauss’s law
ර E. dA =
𝑞
𝜀0
1
‫ ׯ‬E. dA = 𝜀 ‫𝑣𝑑 𝜌 𝑣׬‬
0
By Gauss divergent theorem
ර E. dA = න 𝐷𝑖𝑣 𝑬 𝑑𝑣
𝑣
1
1
න 𝜌 𝑑𝑣 ⇒⇒ න (𝐷𝑖𝑣 𝐸 − 𝜌 )𝑑𝑣
𝜀0 𝑣
𝜀0
𝑣
𝑣
1
(𝐷𝑖𝑣 𝐸 − 𝜀 𝜌 )=0
0
1
𝐷𝑖𝑣 𝐸 = 𝜌
𝜀0
This is the differential form of Gauss's law
Maxwell 1st equation
න 𝐷𝑖𝑣 𝑬 𝑑𝑣 =
Electric Field due to Infinite Line of Charge
• Consider a section of infinite line of charge having uniform linear charge
density λ.
• The Electric field intensity at any point ‘P’ which is at distance r from the
wire. For this we consider cylindrical Gaussian surface which passes
through point P. The electric flux passing through the cylinder is given as
ϕ𝑒 = ර E. dA
• The surface S of the cylinder consist of three parts i.e., S1,S2 and S3, where,
S1= Area of top cross section of cylindrical Gaussian surface
S2= Area of bottom cross section of cylindrical Gaussian surface
S3= Area of curved part of Gaussian surface
φe =‫ 𝑆׬‬E. dA +‫ 𝑆׬‬E. dA +‫ 𝑆׬‬E. dA
1
2
3
Now, ‫ 𝑆׬‬E. dA = ‫ 𝑆׬‬E. dA = 0 (because E is perpendicular to dA or S1 and S2)
1
2
and φe = ‫ 𝑆׬‬E. dA =‫ 𝑆׬‬EdA cos𝜃=‫ 𝑆׬‬E dA (because E is parallel to dA or S3)
3
3
3
ϕe = 𝐸 ‫ 𝑆׬‬dA ----------------------------1
3
φe =E (2πrh) -----------------------------2
by Gauss law
𝑞
φe =
𝜀0
∵ ‫ 𝑆׬‬dA =2πrh
3
-------------------3
As the line of charge has constant linear charge density λ,
Therefore: λ= q/h  q=λh
So, the equation (3) becomes:
𝜆ℎ
φe =
𝜀0
-------------------4
• Comparing Eq. (2) and (4), we get:
E
𝜆ℎ
(2πrh)=
𝜀0
⇒⇒ 𝐸=
𝜆
2πr 𝜀0
• If ‘ r̂ ’ gives the direction of electric field intensity, then
𝜆
𝐸=2πr 𝜀 𝑟�̂
0
This expression gives the electric field intensity due to infinite line of charge
Electric field Due to Infinite Sheet of Charge
• Consider an infinite sheet of charge having constant surface charge
density ‘σ’. The figure shows a small portion of such sheet.
• electric field intensity at point ‘P’ which is at the distance ‘r’ from sheet.
For this we consider a cylindrical Gaussian surface is shown in the figure.
The net electric flux passing through the sheet is given as,
φe =‫ 𝑠׬‬E. dA
• We divide the cylindrical Gaussian surface into three parts i.e., S1,S2 and
S3, where,
S1= Area of top cross section of cylindrical Gaussian surface
S2= Area of bottom cross section of cylindrical Gaussian surface
S3= Area of curved part of Gaussian surface
• Thus,
φe =‫ 𝑆׬‬E. dA +‫ 𝑆׬‬E. dA +‫ 𝑆׬‬E. dA ------------------------------1
1
2
3
In this case surfaces S1 and S2, E||dA are parallel to each other i. e. 𝜃 =0
and |dA1|=|dA2|= |dA|
φe =‫ 𝑆׬‬E dA +‫ 𝑆׬‬EdA
1
2
φe =E ‫ 𝑆׬‬dA +E‫ 𝑆׬‬dA
1
2
φe= EA+ EA= 2EA ----------------2
E=constant
by Gauss law
𝑞
φe =
𝜀0
-------------------3
• surface charge density σ = q/A
• Therefore q= σ A. So, the equation (3) becomes:
σ𝐴
φe = -------------------4
𝜀0
• Comparing Eq. (2) and (4), we get:
σ𝐴
σ
2EA= 𝜀 ⇒⇒ 𝐸=2𝜀
0
0
• If ‘ r’̂ gives the direction of electric field intensity, then
σ
𝐸= 𝑟�̂
2𝜀0
This expression gives the electric field intensity due to infinite sheet of
charge
Electric Field due to Spherical Shell of Charge
• Consider a thin spherical shell of radius ‘R’ which have the charge
‘q ’ with constant surface charge density ‘σ ’. The surface charge density
𝑞
𝑞
• 𝜎 = 𝐴 = 4𝜋𝑟2
∵ 4𝜋𝑟2=Surface area of sphere
• 𝑞 = 𝜎 (4𝜋𝑟2)
• Consider a point ‘ P’ outside the shell. We want to find out electric field
intensity due to this charge distribution. For this we consider a spherical
Gaussian surface of radius r which passes through point ‘P’,
• According to Gauss’s law,
𝑞
ර 𝑬 ∙ 𝒅𝑨 =
𝜀0
𝑞
‫𝜀 = 𝜃𝑠𝑜𝑐 𝐴𝑑𝐸 ׯ‬
E||dA, 𝜃=0
𝑞
𝑞
2
𝐸 ර 𝑑𝐴 = ⇒⇒ 𝐸 4𝜋𝑟 =
𝜀0
𝜀0
𝑞
𝐸=
4𝜋𝜀0𝑟2
Thus the uniform spherical shell of charge behaves like a point charge for all
the points outside the shell.
0
• Case-II: Electrostatic force on a charged particle placed inside the
shell.
Consider a point ‘ P’ inside the shell. We want to find out electric field
intensity ‘E’ at point ‘P’ due to this symmetrical charge distribution. For
this we consider a spherical Gaussian surface of radius which passes
through point ‘r’.
According to Gauss’s law,
𝑞
ර 𝑬 ∙ 𝒅𝑨 =
𝜀0
• Because the Gaussian surface enclose no charge, therefore ‘q = 0’,
ර 𝑬 ∙ 𝒅𝑨 = 0 ⇒⇒ ර 𝐸 𝑑𝐴 = 0
𝐸 ර 𝑑𝐴 = 0
∴ 𝐸 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
As dA ǂ0 therefore E=0
So the electric field does not exist inside a uniform shell of charge. So the
test charge placed inside the charged shell would experience no force.
Electric Field due to Spherical Charge
• Case-I: electric field intensity outside solid sphere of charge.
Consider a spherical distribution of charge of radius ‘R ’ with the uniform
volume charge density ‘ρ’. The electric field at point ‘r’ at a distance from the
center of charged sphere. For we consider a spherical Gaussian surface which
passes through point ‘P ’ as shown in the figure.
According to Gauss’s law,
𝑞
ර 𝑬 ∙ 𝒅𝑨 =
𝜀0
𝑞
ර 𝐸𝑑𝐴 𝑐𝑜𝑠𝜃 =
𝜀0
As the electric field is radial, so the electric lines of force leave the Gaussian surface
normally at all points. Therefore, the electric field intensity E and surface area
element dA are in same direction i.e., .
ර 𝐸𝑑𝐴 =
𝑞
𝜀0
𝑞
𝜀0
𝑞
𝐸 4𝜋𝑟2 =
𝑞 𝜀0
𝐸=
4𝜋𝜀0𝑟2
E ර 𝑑𝐴 =
Case-II: Electric field intensity inside solid sphere of charge.
Consider a spherical distribution of charge of radius ‘ R’ with the uniform volume
charge density ‘ ρ’. The total charge in this uniform charge distribution is
𝑞
𝑞
ρ= =
−−−−− −1
𝑉 4/3𝜋𝑟3
∵ 4/3𝜋𝑟3= Volume of sphere
4
• 𝑞 = 𝜎 (3𝜋𝑟3)
• We want to find the electric field at point ‘P’ at a distance r<R from the center of
charged sphere. For this we consider a spherical Gaussian surface of which passes
through point P as shown in the figure.
Let the Gaussian surface encloses the charge q’<q given by
1
𝑞′ = 𝜎
−−−−− −2
4/3𝜋𝑟3
Dividing eq. (1) and (2), we get
4
𝑞′ 𝜌(3 𝜋𝑟3)
=
𝑞 𝜌(4 𝜋𝑅3)
3
𝑞′ 𝑟3
𝑟3
′
=
⇒⇒ 𝑞 = 3 q
𝑞 𝑅3
𝑅
According to Gauss’s law
𝑞′
𝑞′
ර 𝑬 ∙ 𝒅𝑨 = ⇒⇒ ර 𝐸𝑑𝐴 𝑐𝑜𝑠𝜃 =
𝜀0
𝜀0
E||dA, 𝜃=0
𝑞′
E ර 𝑑𝐴 =
𝜀0
𝑞′
2
𝐸 4𝜋𝑟 =
𝜀0
𝑞′
𝐸=
−−−−−− −3
2
4𝜋𝜀0𝑟
Putting the value in eq. (3), we get
𝑞 𝑟3
𝐸=
4𝜋𝜀0𝑟2 𝑅3
𝑞 𝑟
𝐸=
4𝜋𝜀0 𝑅3
This is expression of electric field intensity inside solid sphere of charge.
Case-III: electric field intensity at the surface of solid sphere of charge
Put r=R
1 𝑞
𝐸=
4𝜋𝜀0 𝑅2
Variation of Electric Field Intensity as a function of
distance for Volume charge distribution
• The graphical representation of the dependence of electric field
strength on the radial distance ‘r’ from the center of this charge
distribution is shown in the figure:
• Electric field intensity, inside solid sphere of charge, is directly
proportional to distance as described by formula:
𝑞 𝑟
𝐸=
4𝜋𝜀0 𝑅3
So the graph between and is a straight line for the values of r from
0R .
• The electric field intensity is maximum at the surface of sphere of
charge:
1 𝑞
𝐸=
4𝜋𝜀0 𝑅2
• The solid sphere of charge behaves as a point charge for all the
points outside the solid sphere of charge i.e., electric field
intensity is inversely proportional to the square of the distance
from center of sphere of charge
Coulomb’s Law from Gauss’s Law
Coulomb’s law can be deduced from Gauss’s law under certain symmetry
consideration. Consider positive point charge ‘q0’. In order to apply the
Gauss’s law, we assume a spherical Gaussian surface as shown in the figure.
Considering the integral form of Gauss’s law,
𝑞
ර 𝑬 ∙ 𝒅𝑨 =
𝜀0
Because the both vectors E and dA are directed radially outward, so
𝑞
E ‫= 𝐴𝑑 ׯ‬
𝜀0
𝑞
2
E 4π𝑟 =
𝜀0
1 𝑞
𝐸=
4𝜋𝜀0 𝑟2
Example-1. A plastic rod whose length is 220 cm and whose radius is 3.6 mm
carries a negative charge q of magnitude -3.8 x10-7 C spread uniformly over
its surface. What is the electric field near the midpoint of the rod at a point
on its surface?
Solution: l=220cm=2.2m
r=3.6 x10-3m
q= -3.8 x10-7 C
E=?
As Electric field intensity due to infinite line of charge is:
𝜆
𝐸=
2𝜋𝜀0𝑟2
𝑞 −3.8 × 10−7
𝐶
−7
𝜆= =
= −1.73 × 10
𝑙
2.2
𝑚
1
(−1.73 × 10−7 )
𝑁
5
𝑁𝑜𝑤 𝐸 =
= −8.6 × 10
−12
−3
2
(2 × 3.14 × 8.85 × 10 ) 3.3 × 10
𝐶
Example. Two equal and opposite charges of magnitude 1.88 x10-7C and
held 15.2 cm apart. What is direction and magnitude of E at midway point
between the charges? What is the force act on an electron placed here?
Solution: q1= 1.88 x10-7C, q2= -1.88 x10-7C
Distance r=15.2cm =15.2 x10-2m
Total electric field | E+| =|E-|=?
Force on an electron placed at the same point F=?
−7
1 𝑞
𝑞1
1.88
×
10
𝑁
9
6
|𝐸 + | =
= 𝑘 2 = 9 × 10
= 0.32 × 10
2
−2
2
4𝜋𝜀0 𝑟
𝑟
15.2 × 10
𝐶
−7
1 𝑞
𝑞2
1.88 × 10
𝑁
9
6
|𝐸 − | =
= 𝑘 2 = 9 × 10
= 0.32 × 10
2
−2
2
4𝜋𝜀0 𝑟
𝑟
15.2 × 10
𝐶
Total electric field E=| E+|+ |E-|=0.32 X106 + 0.32 X106 =0.64 X106 N/C
Now
Force on an electron placed at the same point F=qE= -1.6X10-19 X 0.32 X106
F=-1.024 X10-13N