Up next in 10
Field Due to Continous Distribution of Charges
Jun 10, 2025
This video is a very short explanation about field due to continuous distribution of charges. With introduction to line, surface and volume charge densities
View Video Transcript
0:00
before moving any further and applying
0:03
gau's law we must have some knowledge
0:06
about field due to continuous
0:08
distribution of
0:10
charges we already know about field and
0:13
force due to discrete
0:16
charges to know about continuous
0:19
distribution of charges let us assume
0:21
that charges on a Surface are located
0:24
very close together in such a way that
0:28
such a system of charges can be assumed
0:31
to have continuous distribution of
0:34
charges now in a system of closely
0:37
spaced charges total charge could be
0:40
continuously distributed among some line
0:44
over a surface or throughout a
0:47
volume now let us consider a surface of
0:50
arbitrary shape and size we now divide
0:54
the continuous charge distribution into
0:57
small elements containing charge Delta Q
1:01
amount of charge as shown in this figure
1:04
now electric field at any point a due to
1:08
element carrying charge Delta Q is given
1:10
by this equation which is Delta e = 1
1:14
upon 4 Pilon not Delta Q upon r² R now
1:19
here this R is the distance of element
1:22
under consideration from point A and
1:26
this R cap is the unit Vector in the
1:29
direction ction from charge element
1:31
towards point
1:35
a now total electric field at Point a
1:39
due to all such charge elements in the
1:43
charge distribution is given by this
1:45
equation where this index I refers to
1:48
the highest charge element in the entire
1:51
charge
1:54
distribution now since the charge is
1:57
distributed continuously over some
1:59
region
2:00
the sum becomes
2:02
integral hence total field at a within
2:06
the limit Delta Qi TS to zero is given
2:09
by this
2:11
equation here this integral is done over
2:15
the entire charge
2:18
distribution now if a charge Q is
2:21
uniformly distributed along a line of
2:23
length L the line charge density Lambda
2:28
is defined as lamb LDA = Q by L and unit
2:32
of this Lambda is per met also for
2:36
charge distributed nonuniformly over
2:38
align linear charge density would be
2:41
Lambda is equal to DQ upon DL where this
2:44
DQ is the amount of charge in a small
2:47
length element
2:51
DL now if a charge Q is uniformly
2:55
distributed over surface of area
2:57
a the surface charge density Sigma is
3:01
defined as Sigma is = to Q by a and unit
3:05
of surface charge density is Kum per M
3:08
Square again for nonuniform distribution
3:12
of charges over a
3:14
surface surface charge density would be
3:16
Sigma = DQ upon
3:20
da where this da is a small area element
3:24
of charge
3:25
TQ similarly for uniform charge
3:28
distributions volume charge density is
3:31
given as row is = Q by V and for
3:34
non-uniform distribution of charges row
3:37
is equal to DQ upon DV now unit of
3:40
volume charge density is Kum per M
3:44
Cub for more information please visit
3:48
physics catalyst.com
#Physics
#Primary & Secondary Schooling (K-12)
#Science