Electric field for k12 student

1
ELECTRIC FIELD
A SHORT NOTES
Arun Umrao
https://sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING

2 Electric Field
Contents
1 Electric Field 3
1.1 Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Electric Field (Quantitatively Approach) . . . . . . . . 3
1.1.2 Electric Dipole . . . . . . . . . . . . . . . . . . . . . . 6
1.1.3 Electric Field Due To Electric Dipole . . . . . . . . . . 6
1.1.4 3-Dimensional Electric Field Problems . . . . . . . . . 16
1.2 Numerical Computation . . . . . . . . . . . . . . . . . . . . . 20
1.2.1 Electric Field . . . . . . . . . . . . . . . . . . . . . . . 20
1.2.2 Electric Potential . . . . . . . . . . . . . . . . . . . . . 24
1.3 Displacement Field . . . . . . . . . . . . . . . . . . . . . . . . 26
1.4 Electric Force . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.4.1 Electric Dipole in Electric Field . . . . . . . . . . . . . 27
1.4.2 Oil Drop Experiment . . . . . . . . . . . . . . . . . . . 32
1.5 Charge Density . . . . . . . . . . . . . . . . . . . . . . . . . . 34
1.6 Motion of Charged Particle in Electric Field . . . . . . . . . . 34
1.6.1 Motion of Charge in Uniform Electric Field . . . . . . 34
1.7 Relative Permittivity . . . . . . . . . . . . . . . . . . . . . . . 37
1.8 Electric Field Due To Charge Distribution . . . . . . . . . . . 37
1.8.1 Charged Rod At Axial Position . . . . . . . . . . . . . 38
1.8.2 Charged Rod On Equatorial Position . . . . . . . . . . 39
1.8.3 Charged Rod On Un-Symmetrical Position . . . . . . . 42
1.8.4 Charged Ring At Position On Its Axis . . . . . . . . . 45
1.8.5 Charged Disk On Its Axis . . . . . . . . . . . . . . . . 46
1.8.6 Cavity in a Non-Conducting Sphere . . . . . . . . . . . 49
1.9 Electric Force on Surface of Conductor . . . . . . . . . . . . . 51

1.1. ELECTRIC FIELD 3
1Electric Field
1.1 Electric Field
Electric field is the region around a charged body within which it can exert its
electrostatic influence. Electric field is extends to infinity but it falls rapidly
with distance. Electric field is a vector quantity.
1.1.1 Electric Field (Quantitatively Approach)
A charge q2 is placed at finite distance from charge q1 then they experience an
electrostatic force. Space around charge q1 where other charge q2 experience
electrostatic force is called electric field of charge q1. Now the electric field
of the charge q1 at the point of charge q2 is
~
E =
Fq1q2
q2
(1.1)
Where Fq1q2 is force between two charge bodies.
Space around charge or charged body within which an experimental
unit positive charge is attracted or repulsed by charge or charged body is
called electric field space of charge or charged body. The field which is
responsible for this force is called electric field.

4 Electric Field
Electric Field ~
D
We generally define the electric field for vacuum or air medium and permit-
tivity of air or vacuum is taken as ε0. Now assume that the electric field is
measured in a medium of dielectric constant K and permittivity of ε(kε0),
then the electric field in the medium is
~
E =
1
4πε
Q
r2
(1.2)
Here the quantity 1
4π
Q
r2 is same for all mediums and the electric field varies by
permittivity of the medium. Assuming this quantity as a new electric field
vector, defined by ~
D, then the above equation becomes
~
E =
~
D
ε
(1.3)
Or
~
D = ε ~
E (1.4)
The dimensions of D are QL−2
, and the SI units are Cm−2
. This may seem
to be rather trivial, but it does turn out to be more important than it may
seem at the moment.
Intensity of Electric Field
Assume a unit positive experimental charge, q0, is placed near to another
charge Q. Force experienced by experimental charge is known as electric
field at that point due to charge Q.
~
Fk
q0
= k
Q
r2
or
~
E =
1
4πε0
Q
r2
(1.5)
Direction of electric field is directed away from positive charge and directed
to negative charge.
Solved Problem 1.1 A charged particle having charge of 2.00 × 10−13
C is
placed in an electric field of intensity 1.39 × 10−3
N/C. Find the force on it.

Solution Force on charge placed in external electric field is given by q ~
E.
Hence force on the charge 2.00 × 10−13
C in electric field 1.39 × 10−3
N/C is
F = 2 × 10−13
× 1.39 × 10−3
It gives force on charge of magnitude 2.78 × 10−16
N.
Electric Field Line
Electric field lines originated from positive charge and ended at negative
charge. These lines make open path. A unit positive experimental charge
always follow path on electric field line when it is set free to move.
Characteristics of Electric Field Line
Electric field lines have following characteristics: -
1. Electric field lines generated from positive charge (source) and end
at negative charge (sink). This statement can be redesigned as “the
electric field lines emanate from higher potential source and end at the
lower potential sink”.
2. Electric field lines make open curve.
3. Electric field lines never intersect to each other.
4. Electric field lines are appear to enter into a surface normally.
5. Electric field lines never enter into conductor’s surface.
~
E
(F : 1)
~
E
(F : 2)
~
E
(F : 3)
~
E
(F : 4)
Figure 1.1: Interaction between electric field and conductor’s surfaces.

6 Electric Field
Electric field lines are originated from positive charge and ended at
negative charge.
1.1.2 Electric Dipole
It is a system of two unlike point charges of equal magnitude placed at finite
distance. The line joining to the charges of an electric dipole is known as its
axis and its distance is assumed as 2d or 2l, i.e. multiple of 2.
+q −q
2l
Figure 1.2: Electric dipole.
Electric dipole moment ~
p is p = q2l where 2l is the displacement vector
pointing from the negative charge to the positive charge. SI unit of electric
dipole is C-m. The electric dipole moment vector ~
p points from the negative
charge to the positive charge. An idealization of this two-charge system is the
electrical point dipole consisting of two (infinite) charges only infinitesimally
separated, but with a finite ~
p.
1.1.3 Electric Field Due To Electric Dipole
There are two symmetrical position of point around the electric dipole where
intensity of electric field is to be calculate. These are (i) axial position and
(ii) equatorial position.

Electric Field At Axis of Electric Dipole
+q −q E+q
E−q
bc
C
2l
r
Figure 1.3: Electric field at point due to electric dipole.
The field around charges is
Figure 1.4: Electric field lines at test point charge due to electric dipole.
From figure (1.3) a point C is at distance r from the center of electric
dipole in dipole axis. 2l is length of electric dipole. At point C electric field
intensity due to positive charge of electric dipole is
~
E+q =
1
4πε0
| + q|
(r + l)2
Away from electric dipole. (1.6)
and due to negative charge of electric dipole is
~
E−q =
1
4πε0
| − q|
(r − l)2
Towards electric dipole. (1.7)
Direction of electric fields due to positive charge is away from center of electric
dipole and due to negative charge is towards center of electric dipole. Both
directions are opposite and parallel then resultant electric field due to electric
dipole at point C is equal to | ~
E+q|−| ~
E−q|. Hence resultant electric field from

8 Electric Field
equations (1.6) and (1.7) at point C is
~
E =
1
4πε0
| + q|
(r + l)2
−
1
4πε0
| − q|
(r − l)2
=
1
4πε0

q
(r + l)2
−
q
(r − l)2

=
q
4πε0

(r − l)2
− (r + l)2
(r2 − l2)2

=
q
4πε0

r2
+ l2
− 2rl − r2
− l2
− 2rl
(r2 − l2)2

Or
~
E =
q
4πε0

−4rl
(r2 − l2)2

(1.8)
Equation (1.8) is applicable when 2l is not negligible with respect to distance
r. If l r then l2
r2
so leaving negligible value l2
hence
~
E =
q
4πε0

−4rl
r4

or
~
E =
1
4πε0

−2q.2l
r3

Using p = q.2l where p is electric dipole moment.
~
E =
1
4πε0

−2p
r3

(1.9)
This is equation for electric field at point C due to electric dipole. −p shows
that electric field at point C due to negative charge is greater than electric
field at point C due to positive charge.

At Equatorial Position of Electric Dipole
r E−q
E+q
2l
bc
C
−q
P
θ +q
Q
E+q
E−q
Eh
+q + Eh
−q
Ev
+q
Ev
−q
bc
C
θ
Figure 1.5: Electric field due to electric dipole in equatorial position.
Figure 1.6: Electric field lines at test point charge due to electric dipole.
From first part of figure (1.5), point C is at distance r from the center of
the electric dipole. 2l is length of electric dipole. At point C electric field
intensity due to positive charge of electric dipole is
~
E+q =
1
4πε0
| + q|
(QC)2
(1.10)
and due to negative charge of electric dipole is
~
E−q =
1
4πε0
| − q|
(PC)2
(1.11)
Direction of electric field due to positive charge is from Q to C and due
to negative charge is from C to P. From second part of figure (1.5) ~
E1

10 Electric Field
has two parts, one ~
E1 cos θ parallel to horizontal axis towards Q to P and
~
E1 sin θ vertically upward. Similarly ~
E2 has two parts, one ~
E2 cos θ parallel to
horizontal axis towards Q to P and ~
E2 sin θ vertically downward. Vertically
downward and upward parts are parallel and opposite in direction hence they
will cancel each other. So total Electric field at point C is ~
E = | ~
E1| cos θ +
| ~
E2| cos θ. Substituting values of ~
E1 and ~
E2 the electric field at C will be
~
E =

1
4πε0
+q
(QC)2
+
1
4πε0
+q
(PC)2

cos θ
Where from figure (??), QC = PC =
p
(r2 + l2) and cos θ = l
√
(r2+l2)
. Hence
~
E =
1
4πε0

+q
(r2 + l2)
+
+q
(r2 + l2)

l
p
(r2 + l2)
or
~
E =
1
4πε0

2q l
(r2 + l2)3/2

(1.12)
if l r then l2
r2
so leaving negligible value l2
hence
~
E =
1
4πε0

2q l
(r2)3/2

Using p = q.2l
~
E =
1
4πε0
n p
r3
o
Where p is electric dipole moment.
~
E =
1
4πε0
p
r3

(1.13)
This is equation for electric field at point C due to electric dipole in equatorial
axes. This is half of the electric field at axial point C given by equation (1.9).
Solved Problem 1.2 Two charges 2µC and 3µC are placed at finite distance of
2cm. Find the electric field intensity at distance 10cm in equatorial position
(perpendicular bisector of axis of dipole) of the charge system.
Solution

r
10cm
E2
E3
2cm
bc
C
2µC
P
θ 3µC
Q
E2
E3
Ev
2 + Ev
2
EH
3 EH
2
bc
C
Electric field at the point C due to charges 2µC and 3µC are
E2 =
1
4πε0
×
2 × 10−6
r2
and
E3 =
1
4πε0
×
3 × 10−6
r2
Vector directions of these two electric field lines are away from the two charges
along the line of axis joining charges to the point C. The magnitudes of
electric field intensities E2 and E3 are unequal hence, these field lines should
be resolved in horizontal and vertical components.
E2h =
1
4πε0
×
2 × 10−6
r2
× cos θ
E2v =
1
4πε0
×
2 × 10−6
r2
× sin θ
and
E3h =
1
4πε0
×
3 × 10−6
r2
× cos θ
E3v =
1
4πε0
×
3 × 10−6
r2
× sin θ
Horizontal components will be subtracted while vertical components will be
added. Hence
Eh = −
1
4πε0
×
1 × 10−6
r2
× cos θ
Ev =
1
4πε0
×
5 × 10−6
r2
× sin θ

12 Electric Field
Now r, cos θ and sin θ are given by
r =
√
101 × 10−2
m
cos θ =
1
√
101
sin θ =
10
√
101
Substituting these values in horizontal and vertical components of electric
fields
Eh = −
9 × 107
101
√
101
and
Ev =
450 × 107
101
√
101
Now net electric field is given by
E =
q
E2
h + E2
v
It gives
E = 4.43 × 106
N/C
bc
The direction of this electric field is
tan θ =
Ev
Eh
It gives angle θ = 88.89◦
from negative x-axis.

Solved Problem 1.3 An arc is substanding angle θ at its center. If charge
density in the arc is λ then prove that magnitude of electric field at its center
is
E =
1
4πε0
2λ sin(θ/2)
R
Also find the direction of electric field in your answer.
Solution
θ
E
bc
dl
φ
Eh
A small element dl in the arc is producing electric field E at its center is
given by
E =
1
4πε0
λdl
R2
Due to symmetry of arc, only horizontal component of electric field shall be
effective, hence
Eh = E cos φ
Here φ is instantaneous angular direction of electric field from horizontal.
Now
Eh =
1
4πε0
λdl
R2
cos φ
Again for arc element dl = R dφ and
Eh =
1
4πε0
λdφ
R
cos φ
Integrating it for the angular limit of arc from −θ/2 to θ/2, electric field is
Eh =
1
4πε0
2λ sin(θ/2)
R
Electric field will be in outward direction along the equatorial axis of the arc.

14 Electric Field
Geometrical Approach
1
Two like charges, Q1 and Q2 are placed at points (x1, y1) and (x2, y2)
respectively. As both charges are like, hence neutral point will be in between
these two coordinate points. Let neutral point is at (x, y). A, positive unit
test charge is placed here to get the electric field intensity. If electric field
intensity at this point is zero then horizontal and vertical components of
electric field intensities due to both charges must be equal and in opposite
direction as shown in the figure.
x
y
b
Q1
(x1, y1)
b
Q2
(x2, y2)
b
(x, y)
ExQ1
EyQ1
ExQ2
EyQ2
Electric field at at neutral point due to charge Q1 is
EQ1 =
1
4πε0
Q1
r2
1
(1.14)
Here r1 =
p
(x1 − x)2 + (y1 − y)2. Gradient of these two points (x, y) and
(x1, y1) is
tan θ1 =
y1 − y
x1 − x
Horizontal components of electric field due to charge Q1 is
ExQ1
=
1
4πε0
Q1
r2
1
| cos θ1| (1.15)
Here cos θ1 is
cos θ1 =
|x1 − x|
p
(x1 − x)2 + (y1 − y)2
Similarly vertical component of electric field due to charge Q1 is
EyQ1
=
1
4πε0
Q1
r2
1
| sin θ1| (1.16)
1
Needs Approval

(1.19)
Similarly, equating vertical components, we have
Q1
Q2
=
r2
1
r2
2
×

(1.20)
Comparing above two relations, we get
| tan θ1| = | tanθ2| (1.21)

16 Electric Field
This will gives relation between x and y, if two coordinates are known. Ac-
cording to relation
x1 − x
y1 − y
=
x − x2
y − y2
(1.22)
Now substitute value of x or y in any of two equations 1.14 or 1.20 we will
get the value of x. Now we can calculate value of y from relation 1.22.
1.1.4 3-Dimensional Electric Field Problems
y
z
x
~
r1
~
E1
~
r2
~
E2
~
E
bc
C
bc
A
bc
B ~
E1
~
E2
~
E
bc
Let two point charges ±Q1 and ±Q2 are placed at two points P and Q whose
coordinates are (x1, y1, z1) and (x2, y2, z2) respectively. A positive test charge
is placed at third point C of coordinate (x, y, z) where electric field is to be
calculated. Now, electric field at C due to charge at point P is
~
E1 =
1
4πε0
±Q1
|~
r1|3
~
r1 (1.23)

2
Similarly electric field at C due to charge at point Q is
~
E2 =
1
4πε0
±Q2
|~
r2|3
~
r2 (1.24)
The values of ~
r1 and ~
r2 are
~
r1 = (x − x1)î + (y − y1)ĵ + (z − z1)k̂ (1.25)
and
~
r2 = (x − x2)î + (y − y2)ĵ + (z − z2)k̂ (1.26)
respectively. The magnitude of vectors ~
r1 and ~
r2 are
r1 =
p
(x − x1)2 + (y − y1)2 + (z − z1)2 (1.27)
and
r2 =
p
(x − x2)2 + (y − y2)2 + (z − z2)2 (1.28)
respectively. Net electric field at point C is
~
E = ~
E1 + ~
E2 (1.29)
Substituting the values from equations 1.23 and 1.24 we have
~
E =
1
4πε0
±Q1
|~
r1|3
~
r1 +
1
4πε0
±Q2
|~
r2|3
~
r2 (1.30)
2
We know that the electric field
E =
1
4πε0
Q
r2
is a vector quantity and in vector form it is written as
~
E =
1
4πε0
±Q
r2
r̂
For a given vector, unit vector r̂ is given by
r̂ =
~
r
r
So the vector form of electric field is
~
E =
1
4πε0
Q
r3
~
r

18 Electric Field
On simplification
~
E =
1
4πε0

±Q1
|~
r1|3
~
r1 +
±Q2
|~
r2|3
~
r2

(1.31)
This is the required value of electric field at the point C.
Solved Problem 1.4 A student wants to use the properties of vectors to
measure the electric field intensity at a point C. She takes two positive
point charges of magnitudes 3µC and 4µC respectively and places them at
points A and B respectively. She measures the magnitude of electric field at
the point C. Find the magnitude of electric field at point C. Also find the
direction of electric field. The point coordinates are A=(3, −1, 5), B=(3, 0, 3)
and C=(1, 0, 3).
Solution The vector ~
AC and ~
BC are
~
AC = (1 − 3)î + (0 − (−1))ĵ + (3 − 5)k̂
and
~
BC = (1 − 3)î + 0ĵ + (3 − 3)k̂
And magnitudes of these vectors are | ~
AC| = 3 and | ~
BC| = 2. Now electric
field vector at point C is
~
E =
1
4πε0

3 × 10−6
| ~
AC|3
~
AC +
4 × 10−6
| ~
BC|3
~
BC
#
Substituting the values of vectors and their magnitudes
~
E =
10−6
4πε0

3
27
(−2î + ĵ − 2k̂) +
4
8
(−2î)

On simplification
~
E =
10−6
4πε0

−11
9
î +
1
9
ĵ +
−2
9
k̂

1. Magnitude of the electric field at the point C in unit of N/C is
obtained by finding the magnitude of above equation. Hence
E =
42
3
× 103
2. The direction of electric field at the point is direction of unit vector
of ~
E obtained by using relation ~
E/| ~
E|.

Solved Problem 1.5 Three point charges 2µC, −2µC and −2µC are placed
at coordinates (0, 0), (0, 2) and (3, 0) respectively. Find the electric field
strength and its direction at coordinate point (4, 1).
Solution
0
1
2
0 1 2 3
î
ĵ
A
B
C
O
r1
r2
r3
bc
b
b
bc
(a)
0
1
2
0 1 2 3
~
Ex
~
Ey
A
B
C
O
bc
b
b
EA
EB
EC
bc
(b)
~
Ex
~
Ey
EHA
EHB
EHC
EV A
EV B
EV C
bc
(c)
The vectors ~
AO, ~
BO and ~
CO are
~
AO = 4î + ĵ
~
BO = 4î − ĵ
and
~
CO = î + ĵ
And magnitudes of these vectors are | ~
AO| =
√
17, | ~
BO| =
√
17 and ~
CO| =
√
2. Now electric field vector at point O is
~
E =
2 × 10−6
4πε0

1
17
√
17
(4î + ĵ) +
−1
17
√
17
(4î − ĵ) +
−1
2
√
2
(î + ĵ)

1. On simplification
~
E =
2 × 10−6
4πε0

−1
2
√
2
î +

2
17
√
17
−
1
2
√
2

ĵ

2. The angular direction of electric field with respect to x-axis is
tan θ =
2
17
√
17
− 1
2
√
2
−1
2
√
2

20 Electric Field
Here, numerator and denominator are negative separately, hence the direc-
tion of electric field will be towards third quadrant at point O. Now the
angule value of direction of electric field from x-axis in thrid quadrant is
θ = tan−1

1 −
4
√
2
17
√
17
#
1.2 Numerical Computation
1.2.1 Electric Field
Electric field lines are continuous lines emerge from positive charge and end
at negative charge. Electric field intensity at a point due to a point charge
system is given by
~
E =
1
4πε0εr
q
~
r
|~
r|3
A two charge particles system is shown in given below.
bc
q1
b
q2
b
P(x, y)
~
r1, θ1
~
r2, θ2
A virtual Gaussian surface of equal electric field intensity is drawn around
this charge system. Every point on this Gaussian surface has equal strength
of electric field. A point P(x, y) is chosen on this surface. In the given figure
only a portion of the Gaussian surface is shown.

1.2. NUMERICAL COMPUTATION 21
bc
q1
b
q2
b
P(x, y)
r1, θ1
~
r2, θ2
Electric field on this surface is given by
~
E = ~
E1 + ~
E2
Flux through this surface is given by product of electric field an its area.
Surface area of this portion of the Gaussian surface is not commutable as the
surface may be symmetric or unsymmetrical.
b
q2
b
P(x, y)
~
r2, θ2
S1
S2
~
r
~
r2 − ~
r2 cos θ2
But for small surface area, we can consider it as spherical. The radius of
the circular cross section of this surface is given by
~
r2
= (~
r2 + ~
r2 cos θ2)(~
r2 − ~
r2 cos θ2)
On simplification
~
r2
= ~
r2
2 − ~
r2
2 cos2
θ2
It gives the area of the small surface area
dS1 = π(~
r2
2 − ~
r2
2 cos2
θ2)

22 Electric Field
Now flux in this surface is
dφ = E × dS1
Or
dφ =
1
4πε0εr
×
q2
~
r2
2
× π(~
r2
2 − ~
r2
2 cos2
θ2)
It gives
dφ =
1
4ε0εr
q2(1 − cos2
θ2)
If θ ≈ 0 then (1 − cos2
θ2) = 2(1 − cos θ2). Now
dφ =
1
2ε0εr
q2(1 − cos θ2)
It is clear that flux over the given surface is depend only on the charge and
angular position of point from the charge. Now we can take that
q2(1 − cos θ2) = K
Where K is a constant value. If there are n charge system enclosed by virtual
Gaussian surface then
i=n
X
i=1
qi(1 − cos θi) = K1
Or
i=n
X
i=1
qi cos θi = K2
cos θ for a point charge at placed (x1, y1) with respect point (x, y) is given by
cos θ =
x − x1
p
(x − x1)2 + (y − y1)2
Hence for n point charge system
i=n
X
i=1
qi
x − xi
p
(x − xi)2 + (y − yi)2
= K2
Now we can calculate all possible points for different K2 of electric field lines
for which above relation is satisfied. Now the path of these points is the trace
of electric field line. C program for generation of tracing points of electric
potential surface of two charge system at [1C, −2x, 0y] and [−1C, 2x, 0y] is
given below.

✞
1 #include stdio.h
#include stdlib.h
3 #include math.h
#define bufsize 100
5 char r[bufsize ];
FILE *fp;
7 int n;
9 int main(int argc , char** argv) {
float x, y, k;
11 fp = fopen(a.txt, wb);
for (x = -5; x 5; x = x + 0.1) {
13 for (y = -5; y 5; y = y + 0.1) {
k = (x + 2) / pow((x + 2)*(x + 2) + y*y,
0.5)
15 - (x - 2) / pow((x - 2)*(x - 2) + y*y,
0.5);
if(k 0.2409 k 0.25001) {
17 n = sprintf (r, %2.2ft%2.3ft%2.3fn,
k, x, y);
fwrite(r, sizeof (char), n, fp);
19 }
}
21 }
fclose(fp);
23 return (EXIT_SUCCESS);
}
✌
✆
A tabulted data table is saved in the file ‘a.txt’.
Solved Problem 1.6 Find the tabulated data for electric field lines for charge
system 1C and −0.5C placed at (−1, 0) and (2, 0) respectively.
Solution
✞
#include stdio.h
2 #include stdlib.h
#include math.h
4 #define bufsize 100
char r[bufsize ];
6 FILE *fp;

24 Electric Field
int n;
8
int main(int argc , char** argv) {
10 float x, y, k;
fp = fopen(a.txt, wb);
12 for (x = -12; x = 12; x = x + 0.01) {
for (y = -12; y = 12; y = y + 0.01) {
14 k = (x + 1) / pow((x + 1)*(x + 2) + y*y,
0.5)
- (x - 2) / pow((x - 2)*(x - 2) + y*y,
0.5);
16 if (k 0.295 k 0.30) {
n = sprintf (r, %2.2ft%2.3ft%2.3fn,
k, x, y);
18 fwrite(r, sizeof (char), n, fp);
}
20 }
}
22 fclose(fp);
return (EXIT_SUCCESS);
24 }
✌
✆
1.2.2 Electric Potential
Electric potential surfaces are spherical surface around charges having radii
of potential magnitude at a point on the surface. Electric potential at a point
due to a point charge is given by
V =
1
4πε0εr
q
r
A two charge particles system is shown in given below.
bc
q1
b
q2
b
P(x, y)
r1
r2

At a point P(x, y) electric potential is given by V = V1 + V2 and from
potential definitions:
V =
1
4πε0εr

q1
r1
+
q2
r2

It is clear that potential at a point is depend only on the sum of ratios of
charge to distance.
q1
r1
+
q2
r2
= K
Where K is a constant value. If there are n charges in a system then
i=n
X
i=1
qi
ri
= K1
Here
ri =
p
(x − xi)2 + (y − yi)2
Hence for n point charge system
i=n
X
i=1
qi
p
(x − xi)2 + (y − yi)2
= K1
Now we can calculate all possible points for different K1 of electric poten-
tial surfaces for which above relation is satisfied. Now the path of these
points is the trace of electric field line. C program for generation of tracing
points of electric potential surface of two charge system at [1C, −2x, 0y] and
[−1C, 2x, 0y] is given below.
✞
#include stdio.h
2 #include stdlib.h
#include math.h
4 #define bufsize 100
char r[bufsize ];
6 FILE *fp;
int n;
8
int main(int argc , char** argv) {
10 float x, y, k;
fp = fopen(a.txt, wb);
12 for (x = -5; x 5; x = x + 0.1) {
for (y = -5; y 5; y = y + 0.1) {

26 Electric Field
14 k = 1 / pow((x + 2)*(x + 2) + y*y, 0.5)
- 1 / pow((x - 2)*(x - 2) + y*y, 0.5);
16 if (k -1 k 1) {
n = sprintf (r, %2.2ft%2.3ft%2.3fn,
k, x, y);
18 fwrite(r, sizeof (char), n, fp);
}
20 }
}
22 fclose(fp);
return (EXIT_SUCCESS);
24 }
✌
✆
A tabulted data table is saved in the file ‘a.txt’.
1.3 Displacement Field
Consider an infinite parallel plate capacitor placed in space (or in a medium)
with no free charges present except on the capacitor. In SI units, the charge
density on the plates is equal to the value of the ~
D field between the plates.
I
A
~
D · d ~
A = Qfree (1.32)
−q
+q
~
D
Figure 1.7: Electric field in vacuum.
Outside the capacitor, the fields of the two plates cancel each other and
| ~
E| = | ~
D| = 0.

1.4. ELECTRIC FORCE 27
~
D
~
E
~
D
−q
+q
Figure 1.8: Electric field in presence of material.
If the space between the capacitor plates is filled with a linear homo-
geneous isotropic dielectric with permittivity ε, the total electric field ~
E
between the plates will be smaller than ~
D by a factor of ~
ε, i.e.
| ~
E| =
Qfree
εA
(1.33)
Hence, we can say that ~
D is inducing electric field to a material and ~
E is
induced electric field in the material.
1.4 Electric Force
If a point charge is placed in the electric field of other charged body then they
mutually interact with each other. The strength or intensity of interaction
between them is known as the electric force. Suppose an experimental charge
of +qC is placed in the field intensity ~
E then electric force on the experimental
charge is
F = q ~
E (1.34)
1.4.1 Electric Dipole in Electric Field
External electric field exerts electric force on electric dipole when it is placed
in field. Positive charge on electric dipole experience force in the direction of
electric field and negative charge experience force in the opposite direction of
electric field. Both forces on positive and negative charge made force couple.
These forces are parallel and opposite in direction hence they tend to align
electric dipole parallel to electric field.

28 Electric Field
F−q
F+q
P
Q
~
E
θ
2l
2l cos θ
2l
sin
θ
Figure 1.9: Electric dipole in homogeneous electric field.
Electric field lines
Figure 1.10: Effect of external electric field on electric dipole charges.
Solved Problem 1.7 Two point charges 2µC and 2.5µC are placed by a
distance of 1.2 × 10−1
.m Find the point where electric field intensity is zero.
Solution
2µC 2.5µC
E2
E2.5
bc
C
0.12m
x
bc
Both charges are positive, hence electric field will be zero inside the
charges on the line of axis joining the charges. For this case assume the

null point is at distance of x from 2µC. Now for the null points
2 × 10−6
x2
=
2.5 × 10−6
(0.12 − x)2
On solving it x = 0.0631m right from the 2µC charge.
Solved Problem 1.8 Two point charges 1.2µC and −3.25µC are placed by
a distance of 12cm. Find the point from positive charge where electric field
intensity is zero.
Solution
1.2µC −3.25µC
E1.2 E−3.25
bc
C
0.12m
x
bc
Both charges are unlike charges, hence electric field will be zero outside
the charges on the line of axis joining the charges. For this case assume the
null point is at distance of x from 1.2µC. Now for the null points
1.2 × 10−6
x2
=
3.25 × 10−6
(0.12 + x)2
On solving it x = 0.18m left from the 1.2µC charge.
Electric dipole Moment in Homogeneous Electric Field
From figure (??) an electric dipole is placed in external electric field inclined
at θ angle with electric field. Force on positive charge is
F+q = q ~
E
along the direction of electric field and force on negative charge is
F−q = q ~
E

30 Electric Field
opposite to the direction of electric field. These forces make force couple
hence force moment on electric dipole is
τ = q ~
E 2l sin θ
Where 2l sin θ is perpendicular distance between direction of two forces F+q
and F−q. In terms of electric dipole moment p = q2l,
τ = p ~
E sin θ
To find the direction of electric dipole moment, we place electric dipole
in an external electric field freely. Electric dipole aligns itself along the
direction of electric field. Now the direction of electric field is parallel to
the direction of electric dipole moment. Generally its direction is from more
negative charge to less negative charge.
Electric dipole Moment in Inhomogeneous Electric Field
δx
P Q
~
E ~
E + ~
δE
Figure 1.11: Electric dipole in inhomogeneous electric field.
Consider a simple dipole consisting two charges +q and −q separated by
small distance δx. The force on the charge −q is q ~
E in opposite direction of
electric field and force on the charge +q is q( ~
E + ~
δE) along the direction of
electric field. If p is electric dipole moment then there is a net force to the
right of magnitude q ~
δE. Or
Force = p
~
dE
dx
(1.35)
If potential drop between two point P and Q is V volt then electric field
along the direction of x − axis is
Ex = −
∂V
∂x
(1.36)

This is also known as relation between electric field and electric potential.
Work Done When Electric Dipole Is In Electric Field
With reference from figure (1.12), Force on electric dipole when it is placed
in external electric field is τ = p ~
E sin α. Work required to rotate it a small
angle dα is dW hence
dW = p ~
E sin α dα
on integrating above equation within limit α = θ1 to α = θ2 total work is
W = −p ~
E [cos α]θ2
θ1
~
E → ~
p
bcb θ
dθ
Figure 1.12: Work done in rotating electric dipole in an electric field.
Applying the limits
W = −p ~
E [cos θ2 − cos θ1]
W will be maximum when θ2 = 180◦
→ cos θ = −1 and θ1 = 0◦
→ cos θ = 1.
Now work done is
W = −p ~
E(−1 − 1) = 2p ~
E
Solved Problem 1.9 An electric dipole with dipole moment 4 × 10−9
C-m
is aligned at 30◦
with the direction of a uniform electric field of magnitude
5 × 104
N/C. Calculate the magnitude of the torque acting on the electric
dipole.
Solution Torque on a dipole is given by
τ = pE sin θ

32 Electric Field
Placing the values of parameters
τ = 4 × 10−9
× 5 × 104
× sin 30◦
It gives τ = 104
Nm.
1.4.2 Oil Drop Experiment
The Oil drop experiment was carried out by Robert Millikan and Harvey
Fletcher in 1909 to measure the elementary electric charge.
Apparatus It consists two plates placed one over other. These plates are
connected with regulated voltage source. Electrically charged oil drops are
inserted between the plates and let them fall. Electric field between two plates
is adjusted as the electrostatic force on the charged drop in vertically upward
direction is balanced by the gravitational force on the drop in vertically
downward direction. Originally the whole system was in air, but it can be
placed in vacuum chamber. The motion of oil drop is observed by microscope.
~
E
Fe
Fg
bb
(a)
~
E
Fe
Fg
Fd Fd
bb
(b)
Principle This experiment uses the relation form vacuum chamber
Fg = Fe (1.37)
and for air chamber
Fe = w + Fd (1.38)
Where w is the apparent weight and Fd is drag force.
Measurement In Original Experiment The drop is allowed to fall in the
absence of an electric field. If v is the terminal velocity then
Fd = 6πrηv (1.39)

The oil drop is floating in air medium and assumed a perfect sphere hence
its apparent weight is
~
w =
4π
3
r3
(ρ − ρair)~
g (1.40)
At terminal velocity, drop is not accelerating, hence net force on it is zero
and equation (1.39) and (1.40) are equal. This give the radius of the drop.
r2
=
9ηv
2g(ρ − ρair)
(1.41)
Now electric field is applied and it is adjusted as it applied electrostatic force
on the charged oil drop in vertically upward direction then
Fe = q ~
E (1.42)
If apparent weight is balanced by electrostatic force then Fe = w and from
equations (1.39) and (1.42)
q ~
E =
4π
3
r3
(ρ − ρair)~
g (1.43)
But mass of oil drop can not be measured easily, hence electric field is ad-
justed as the drop starts moving in upward direction and attained a constant
terminal velocity vt, so Fe − w = Fd and
q ~
E −
4π
3
r3
(ρ − ρair)~
g = 6πη(~
r · ~
vt) (1.44)
If two plates are separated by ~
d then electric field and electric potential
relation
V = ~
E · ~
d (1.45)
and relation (1.44) becomes
q

V
~
d

−
4π
3
r3
(ρ − ρair)~
g = 6πη(~
r · ~
vt) (1.46)
This relation can be used to find the charge of oil drop.
Measurement In Vacuum Experiment In vacuum chamber oil drop ex-
periment, there is no drag forces, buoyancy forces etc. Hence for the vacuum

34 Electric Field
chamber oil drop experiment electrostatic force is balanced by gravitational
force. So
Fe = Fg (1.47)
And
q
V
~
d
= mg (1.48)
This relation can be used to find the charge on oil drop.
1.5 Charge Density
Electric field due to any point is E ∝ q
r2 . Electric charge distribution is
of different type, Linear charge distribution, surface charge distribution and
volume charge distribution.
Linear Charge Distribution Charge Q is distributed over a length of ∆l
then linear charge distribution is ρ = Q
∆l
.
Surface Charge Distribution Charge Q is distributed over a surface area
of ∆s then surface charge distribution is ρ = Q
∆s
.
Volume Charge Distribution Charge Q is distributed over a volume of
∆V then volume charge distribution is ρ = Q
∆V
.
1.6 Motion of Charged Particle in Electric
Field
Moving charge produces both magnetic filed as well as electric field. When a
charge enters into the external electric field it interacts with it and produce
mutual interacting electric force. The motion of charge particle may be in
uniform electric field or in non-uniform electric field.
1.6.1 Motion of Charge in Uniform Electric Field
A charge of mass m and having charge q0 enters into external electric field
with parallel to it, at normal and at any angle θ.

1.6. MOTION OF CHARGED PARTICLE IN ELECTRIC FIELD 35
Parallel to Electric Field
When a charge enters parallel to the external electric field an electric force
exerted on the charge. The value of electric force is q0E. If there are no
other force acting on the charge then this force is responsible for acceleration
of the charge particle. From Newton’s Law of motion F = q0
~
E
ma = q0
~
E
a = q0
~
Em (1.49)
~
F
~
E
Figure 1.13: Force on charged particle by external electric field. Charge is
travelling parallely to external electric field.
If ~
E is uniform (that is, constant in magnitude and direction), then the
acceleration of charge particle is constant. The direction of acceleration of
charge depends on its nature. If charge is positive then it will accelerate
along the direction of external electric field. If charge is negative then it will
accelerate in the opposite direction of the external electric field.
Normal To Electric Field
Let a charged particle of mass m and charge q0 is entering into a field at an
angle 90◦
with constant velocity v î along x axis. The direction of electric
field is along y axis and is Eĵ. The acceleration on the charge particle is
along the direction of electric field.

36 Electric Field
~
F
vî
Eĵ
y
x
Figure 1.14: Force on charged particle by external electric field. Charge
entering into electric field normally.
Let the width of electric field is x in which the charge particle experience
the force and it displaced by y along y axis. Initially the velocity of charged
particle along x axis is ~
vî and acceleration is zero. After entering into the
electric field, acceleration is a and is given by equation (1.49)
aĵ =
eq
m
ĵ
This acceleration displaced charged particle by y throughout the flight, hence
d2
y
dt2
ĵ =
eq
m
ĵ
Here t is the time for which charge particle flights through the external
electric field. On integrating above relation velocity of charge particle along
y axis becomes
dy
dt
=
eq
m
t (1.50)
Integrating again above relation (1.50) the vertical displacement y is
y =
eq
m
t2
2
(1.51)
The horizontal displacement x is given by x = vî × t, where time t is given
by relation (1.51)
x = vî × t (1.52)

1.7. RELATIVE PERMITTIVITY 37
1.7 Relative Permittivity
The force between two charges, given by coulomb’s law
Fv =
1
4πε0
q1q2
r2
is measured when the medium between these two charges is air or vacuum.
If the air or vacuum is replaced by another medium, the force between these
two charged reduces. This is because the electric field intensity between
two charges reduced due to reverse polarity in the medium, which opposes
external electric field between charges. If the magnitude of charges and
distance between them is not change then coulomb’s force is
Fm =
1
4πε
q1q2
r2
Here ε is the permittivity of the medium. Permittivity of a medium is the
property of the medium which determines limit of force between
two charges placed in that medium. The ratio of Fv
Fm
is
εr =
1
4πε0
q1q2
r2
1
4πε
q1q2
r2
εr =
ε
ε0
Here εr is the relative permittivity of the medium with respect to air or
vacuum. Thus Relative permittivity is the ratio of coulomb force be-
tween two charges placed in vacuum to the coulomb force between
these two charges, placed in a medium when their magnitude and
distance between them is not changes.
1.8 Electric Field Due To Charge Distribu-
tion
The previous relation are obtained by assuming that both experimental
charge and active charge are taken as point charges. But in practice ac-
tual charge bodies are not in point shape but they are in linear, surface or
volume shapes. Following sections are illustrating the methods of finding
electric field due to different shapes of bodies.

Electric field for k12 student

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (19)

Ähnlich wie Electric field for k12 student

Ähnlich wie Electric field for k12 student (20)

Mehr von Arun Umrao

Mehr von Arun Umrao (20)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

Electric field for k12 student