Question
Question: A parallel plate capacitor with air as the medium between the plates has a capacitance of\(10\mu F\)...
A parallel plate capacitor with air as the medium between the plates has a capacitance of10μF . The area of the capacitor is divided into two equal halves and filled with two media having dielectric constant k1=2 andk2=4. The capacitance will now be
A. 10μFB. 20μFC. 30μFD. 40μF
Solution
As the initial capacitor is divided by two equal parts by area and filled with a medium having dielectric constant k1=2 andk2=4, so the initial capacitor is replaced by two capacitors connected in parallel having same distance between the plates but area of cross-section of the plate is halved. Use the formula for a combination of capacitors in parallel to get the capacitance when the dielectrics are introduced inside the plates.
Formula used:
Capacitance of parallel plate capacitor is given by
C=dϵA, Where
ϵ= permittivity of medium inside the plates of capacitorA=Area of the platesd=distance between the plates
Equivalent capacitance of two capacitor having capacitance C1and C2connected in parallel is
C=C1+C2
Dielectric constant k of any medium is defined as the ratio of its permittivity ϵ to the permittivity of vacuum ϵ0, i.e.
k=ϵ0ϵ
So permittivity of any medium is ϵ=kϵ0
Complete step by step answer:
Capacitance of a parallel plate capacitor having plate area A and distance between the plate d and permittivity of medium between the plate ϵ is given by
C=dϵA
If the medium between the plate is air then ϵ=ϵ0
So C=dϵ0A
Given C=10μF.
Now the area of the capacitor is divided into two equal halves and filled with two media having dielectric constant k1=2 andk2=4.
So now the initial capacitor is now a combination of two parallel plate capacitors connected in parallel having plate area2A.
Let the upper half is filled with a dielectric medium having dielectric constant k1=2 and the lower half is filled with a dielectric medium having dielectric constant k2=4.
The upper half is now a parallel plate capacitor with plate area A1 and distance between the plates is dand having a dielectric medium in between them with dielectric constant k1.
Its capacitance is given by
C1=dϵ1A1=dk1ϵ0A1 (∵ϵ1=k1ϵ0)
Similarly the lower half is now a parallel plate capacitor with plate area A2 and distance between the plates is d and having dielectric medium in between them with dielectric constant k2.
Its capacitance is given by
C2=dϵ2A2=dk2ϵ0A2 (∵ϵ2=k2ϵ0)
Equivalent capacitance of parallel combination of C1 and C2 is given by
Cp=C1+C2=dk1ϵ0A1+dk2ϵ0A2=dϵ0(k1A1+k2A2)
Put the value of k1=2,k2=4,A1=A2=2A
=dϵ0[2×(2A)+4×(2A)]=dϵ0A(1+2)=d3ϵ0A=3C
Where
C=dϵ0A is capacitor of initial plate
Given C=10μF
So
Cp=3×C=3×10μF=30μF
So the correct option is C. 30μF
Additional Information
Capacitance is defined as the ratio of the change of electric charge of a system to the change in electric potential. i.e.
C=ΔVΔq
The unit of capacitance is Farad written as F.
The capacitance of a capacitor is one Farad when one coulomb of charge changes the potential between the plates by one Volt.
Capacitance is also defined as the ability of a circuit to collect and store energy.
Note:
In a problem like this 1st see whether all the units are given in S.I or not. If not S.I then convert to S.I and then calculate.
Also note that the capacitance value always expressed in micro Farad or nano Farad or picoFarad, because its value in S.I units is very small.1μF=10−6F , 1nF=10−9F and 1pF=10−12F
S.I base unit of Farad is F = A2 s4 kg−1 m−2
A=ampere and s=second,
Dimensional formula [F]=[M−1 L−2 T4 A2]