L13 : Numericals - Capacitor & Capacitance, Physics, Class 12 Video Lecture

hello friends this video on capacitors
and capacitance part 13 is brought to
you by exam few calm no more fear from
exam please make sure that you have
watched all the videos till part 12
before going ahead with part 13 so let
us look at problem 1 it asks a slab of
material of dielectric constant K has
the same area as the plates of a
parallel plate capacitor okay so we have
a parallel plate capacitor it says that
this parallel plate capacitor has a slab
of dielectric constant K which has the
same area as that of the plates that
means if the plates have area a this
slab also has area a but has a thickness
three by fourth the distance between the
separation of the plates that means if
this distance is d then this distance is
3 by 4 d okay how is the capacitance
changed when the slab is inserted
between the plates okay
so initially before the slab is
introduced the capacitance is equal to a
epsilon naught by D now let us suppose
the dielectric slab is introduced so
after the dielectric slab is introduced
the new capacitance becomes equal to a
epsilon naught divided by D minus T plus
T by K right so this can be written as a
epsilon naught divided by D minus T is
the thickness of the dielectric slab
which is equal to 3 by 4 D plus 3 by 4 D
divided by K so this becomes equal to a
epsilon naught divided by D plus 3 D so
this becomes equal to D by 4 plus 3 D by
4 T right now this can be written as let
us calculate it on the side now this can
be written as a epsilon naught divided
by we
take be by for comment so this is one
plus three by K so this in turn can
again be written as four a epsilon
nought divided by D into 1 plus 3 by K
now we have to give the value of this
capacitance only in terms of the values
which are given in the problem that is
it has to be given in terms of K and
three by four D so we can say that this
capacitance becomes equal to what is
this a epsilon naught by D epsilon
naught by D is nothing but the old
capacitance so this can be written as 4
C naught divided by 1 plus 3 by K now
this can be written as 4 C naught
divided by K plus 3 into K therefore the
new capacitance C becomes equal to 4
into K divided by K plus 3 times the
previous capacitance so it looks simple
right okay now let us look at the next
problem it says a dielectric slab of
thickness 1 centimeters and dielectric
constant 5 is placed between the plates
of a parallel plate capacitor of plate
alien 0.01 meter square and separation 2
centimeters so what are the values that
are given in this problem it says
dielectric slab of thickness 1
centimeters of dielectric constant K is
equal to find pls between the plates of
a capacitor of plate area ay is equal to
0.01 meter square and separation that is
D is equal to 2 centimeters that is 2
into 10 to the power minus 2 meters this
is also 1 into 10 to the power minus 2
meters now we have to calculate the
change in capacity on introduction or
dial electric so first we will calculate
the capacitance without introduction of
dielectric then we will calculate the
capacitance on introduction of the
dielectric and we will then find out the
change so let us suppose this is my knee
parallel plate capacitor and a
dielectric slab is introduced to this
way so we already know what happens when
a dielectric slab is introduced so let
us first consider before what happens in
the before scenario before any slab is
introduced your capacitance is equal to
ay epsilon not by B right so what is a
is 0.01 epsilon naught is 8 point 8 5
into 10 to the power minus 12 and D is 2
into 10 to the power minus 2 so this
value comes out to be 4 point 4 to 5
into 10 to the power minus 12 farad's so
this is the capacity of the capacitor
when no slab is introduced now let us
suppose that a dielectric slab is
introduced so dielectric slab is
introduced
so if dielectric slab is introduced what
will be the capacitance capacitance will
be given by a epsilon naught divided by
D minus T plus T by 3 C these form you
might be thinking that it is so
difficult to remember so many formulae
even if you don't remember the formula
you can solve it the way we derived this
formula you have to calculate the
potential difference between the plates
and then put the values and find out but
to make things simple and easier many
times it becomes worthy to remember the
formulas for example in this case if you
want if if all the Furze have understood
the logic how to solve it you can
directly solve it like that but then
that will take or that will consume more
of your time so this will be a is
nothing but 0.01 epsilon naught is again
8 point 8 5 into 10 to the power minus
12 divided by B is 2 into 10 to the
power minus 2 minus T is 1 into 10 to
the power minus 2 plus
into ten to the power minus 2 divided by
K that is fine now when you calculate
this you get the value as seven point
three seven five into ten to the power
and minus 12 farad's so this is the
capacitance of the capacitor when the
dielectric slab is introduced therefore
the change in capacitance is equal to C
minus C naught so that becomes seven
point three seven five minus four point
four two five into ten to the power
minus twelve
Ferentz so this comes out to be two
point nine five into ten to the power
minus twelve balance so this is the
change in capacitance on introduction of
the dielectric now the second half of
the question asks what would be the
change if the dielectric slab were
conducting so instead of dielectric slab
if a conducting slab were introduced
then what would have been the change now
we have already calculated the value of
the capacitance before any slab is
introduced that is equal to this now let
us calculate the capacitance if a
conducting slab is introduced so if a
conducting slab of the same thickness
was introduced then this capacitance
would have been equal to a epsilon
naught divided by D minus T where T is
the thickness of the conducting slab so
a will be 0.01 into eight point eight
five into ten to the power minus 12
divided by D is equal to 2 into 10 to
the power minus two minus T is 1 into 10
to the power minus two so this value
comes out to be eight point eight 5 into
10 to the power minus 12 fathoms so this
is the value of capacitance if a
conducting slab is in
between the plates so in this case God
would have been the change in
capacitance the change in capacitance
will be given by C dash minus C naught
so that is 8 point 8 5 into 10 to the
power minus 12 minus 4 point 4 2 5 into
10 to the power and minus 12 so this
comes out to be 4 point 4 2 5 into 10 to
the power minus 12 farad's right so this
is how we solve problems related to
capacitors thank you please visit exam
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