Electric Field
Jun 10, 2025
Introduction to electric field and how to calculate electric field using Coulomb's law
View Video Transcript
0:00
in the previous tutorial we talked about
0:02
electrostatic force coolum's law and the
0:05
principle of superposition and in this
0:08
tutorial we talk about electrostatic
0:10
field electrical interactions between
0:13
charged particles can be reformulated
0:16
using the concept of electric field to
0:19
understand the concept of electric field
0:21
consider that we have two positively
0:23
charged bodies namely A and B and we
0:30
also know that these two positively
0:32
charged bodies will reppel each other
0:34
with a force
0:39
[Music]
0:46
F if we now remove this charge
0:50
B from its position and label its
0:54
position as
0:57
Point p
1:01
then this charged body a is set to
1:04
produce an electric field at this point
1:07
P now this charge a not only produces
1:11
field at this point P but it also
1:15
produces field at all other points in
1:18
its
1:19
vinity so if we place a Charged body B
1:23
either positive or negative at this
1:25
point p a force is exerted on this body
1:28
B by the field produced by body a and
1:32
since electric field exists at all point
1:35
in the region around a this charged body
1:38
B will experience a force at any point
1:42
in the space around body
1:45
a suppose we have to test the presence
1:48
of electric field at a point say p in
1:51
the region around this charged body a
1:55
now to do this we simply place a small
1:58
charg body at this point
2:01
p and if this test charge experiences
2:05
any force of electric origin then
2:08
electric field e is set to exist at this
2:11
point P so we can now say that an
2:15
electric field is set to exist at a
2:18
point if an electric force is exerted on
2:22
a t charge at that point now electric
2:28
field
2:33
like electric
2:37
[Music]
2:39
force is also a vector
2:47
quantity so the properties of electric
2:50
field are determined when both magnitude
2:53
and direction of an electric force are
2:56
specified now
2:58
if f is the force acting on test
3:04
charge Q D placed at a point in an
3:08
electric field then electric field at
3:11
that point would be equal
3:14
to e =
3:17
to F upon Q
3:21
D and
3:23
[Music]
3:26
since f is equal to
3:32
q- e direction of electric field would
3:35
be same as the direction of electric
3:38
force
3:39
now SI unit of electric field is Newton
3:45
per
3:46
Kum so we now know about electric field
3:50
but we do not know how to calculate this
3:52
electric field the electric field at any
3:55
point can be calculated using kum's law
3:58
if we know about both magnitude and
4:02
position of all the charges contributing
4:04
to the
4:07
field now to find the magnitude of
4:10
electric field at a point P at a
4:13
distance R from the charge Q imagine a
4:17
charge q- placed at this point p and
4:21
from kum's law we
4:26
have Force f is equal to K
4:30
q q d upon r² where we know that this
4:37
proportionality constant K is = 1 upon 4
4:41
Pi
4:45
Epsilon and electric field at this point
4:49
p is given
4:51
[Music]
4:55
by e = f upon
5:02
Q D is = to 1
5:06
upon 4 Pi
5:13
Epson Q upon r² now direction of this
5:18
electric field e is away from the charge
5:21
Q if it is positive and it would be
5:25
towards the charge Q if the charge is
5:28
negative now electric field in terms of
5:31
unit
5:32
Vector R cap directed along the line
5:36
from charge Q to point P would
5:40
be a = to
5:44
k q r
5:47
cap upon
5:49
r² where R is the distance from charge Q
5:52
to point p and this equation holds for
5:55
both positive and negative
5:58
charges
6:01
now if there are n number of charges
6:03
present in a system say
6:06
q1 Q2 Etc which are at a distance
6:12
R1 R2 Etc from a given point
6:17
P then each one of the charges exert a
6:21
force on test charge q- placed at this
6:24
point P now from principle of
6:27
superposition net force on the test
6:30
charge q- would
6:33
[Music]
6:35
be F = to
6:39
fub1 plus
6:43
[Music]
6:45
FS2 and so on where this fub1 is forc on
6:49
test charge due to charge q1 and so on
6:54
and the resultant electric field would
6:56
also be the vector sum of the individual
6:58
electric fields
6:59
[Music]
7:01
[Applause]
7:02
so resultant electric field E equals
7:06
to F upon
7:11
q- is equal to
7:15
E1
7:18
plus
7:21
E2 and so on this is also a direct
7:25
result from the principle of
7:27
superposition similar to what we have
7:29
discussed while studying electric force
7:32
on a single charge due to a system of
7:34
multiple
7:36
charges and for more information visit
7:39
physics catalyst.com
7:40
[Music]
#Physics
#Primary & Secondary Schooling (K-12)