Multiple Choice Questions on Capacitors and capacitance

For parallel plate air capacitor
C = (ε₀A)/d
Which clearly shows that capacitance does not depend on charge on both the conducting plates. Hence answer is
(a), (c) and (d)

Equivalent capacitance of two capacitors each having capacitance C are connected in series combination
$\frac {1}{C^'} =\frac {1}{C} + \frac {1}{C} =\frac {2}{C}$
Or,
$C^' = \frac {C}{2}$
Now circuit given in question takes the form

All the three capacitors in above figure are connected in parallel combination
Thus,
C”=C/2 +C/2 + C = 2C
This is the required equivalent capacitance.
Hence (b) is the correct choice

Given that C₀ and Q₀ are the capacitance and charge of the capacitor before the introduction of the dielectric. Now in the presence of the dielectric capacitance of capacitor becomes
C = (εA)/d
But we know that relative permittivity of any material is given by
ε=Kε₀
Where K is dielectric constant of the material.
Thus,
C = (Kε₀A)/d
   = K C₀
Or, C= 2 C₀
Again charge on the capacitor after the introduction of dielectric is given as
Q=CV                                                 (a)
And before the introduction of dielectric
Q₀= C₀V
Now putting the value of C in equation (a) we find
Q=2 C₀V
Or, Q=2 Q₀
Where V is same before and after the introduction of dielectric as voltage source used is a constant voltage source.
Hence (d) is the correct choice

Charge accumulated on the first capacitor Q₁=CV and charge accumulated on second capacitor is Q_2­= (2C) x (2V) = 4CV. Since the capacitors are connected in parallel such that the plates of opposite polarity are connected together, the common potential of the whole system is
$V^' = \frac {Q_2 -Q_1}{C_1 + C_2} = \frac {4CV -CV}{C+2C} =V$
Equivalent capacitance C’=C+2C=3C. hence the final energy of the configuration is
U’=C’V’²/2 = 3CV²/22C)x(2V capacitor is Qe first capacitor
Hence (b) is the correct choice

The potential difference between the plates remains unchanged after the introduction of dielectric because capacitor remains connected to the battery. Thus V remains equal to V₀. Introduction of dielectric increases the capacitance C and hence charge stored in the capacitor also increases since Q=CV thus Q>Q₀. Since plate distance and V remains unchanged this implies Electric field E=V/d also remains unchanged. Energy stored in the capacitor increases as it depends on the capacitance C of the capacitor since U=CV²/2. Thus U>U₀.
Hence (a) and (d) is the correct choice<

Equivalent capacitance of two capacitors each having capacitance C are connected in series combination  $\frac {1}{C^'} =\frac {1}{C_1} +  \frac {1}{C_2}$

from this it is clear that the effective capacitance of a capacitor is reduced when capacitors are connected in series combination.
Hence (a) is the correct choice

Now capacitance of capacitor before inserting metal plate is
C = (ε₀A)/d
And after the insertion of any dielectric plate filling the capacitor partially is
C = (ε₀A)/[(d-t)+t/K]
We know that K for a metal is infinite thus
C = (ε₀A)/(d-t)
Here t=d/2 thus
C’=(2ε₀A)/d=2C
C:C’=1:2
Hence (a) is the correct choice

When the dielectric slab of thickness t and dielectric constant K is inserted between the plates of the capacitor, the potential difference between the plates of the capacitor is given by
V=E[d-t-(t/K)]
In order to maintain the same value of V separation between the plates should be increased by d’ given by
d’=t(1-1/K) or K=t/(t-d’) = 2mm/(2mm-1.6mm) = 5mm
Hence (b) is the correct choice

As clear from the figure capacitors C₁ and series combination of  C₂ , C₃ and C₄ are connected in parallel. If C’ is the capacitance of series combination then,
$\frac {1}{C^"} = \frac {1}{C_2} + \frac {1}{C_3} + \frac {1}{C_4} =   \frac {1}{3} +  \frac {1}{4} +  \frac {1}{5} = \frac {47}{60}$
or
$C_" =\frac {60}{47} \mu F$
Hence equivalent capacitance between A and B is
C=C₁+C’=(2+60/47)μF=3.27 μF
Hence (d) is the correct choice

We have found earlier that equivalent capacitance between points A and B is C=3.27 μF. Total charge stored would be
Q=CV=3.27 μF x 4Volt = 13.8μC
Hence (a) is the correct choice

Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal,
$Q_2=-Q_3$.
Now PD inside the capacitor is given by
$= \frac {Q_2}{C} = \frac {2Q_2}{C} = \frac { Q_2 + Q_2}{2C}= \frac {Q_2 -Q_3}{2C}$
Hence (c) is the correct option

Total charge =$(2C)(2V)+(C)(-V)=3CV$
Common potential =3CV/3C=V
Energy =$\frac {1}{2}(3C)(V)^2=\frac {3CV^2}{2}$
Hence (a) is the correct option

Potential difference across lower and middle branch of circuit is equal to the potential difference across capacitor of upper branch of circuit.
Current flows through 2 Ω resistance from left to right, is given by
$I=\frac {E}{R+r}=\frac {2.5}{2+ .5}=1A$
Therefore
The potential difference across 2Ω resistance
$V=IR= 1 \times 2 = 2V$
Hence potential difference across capacitor is also 2 V.
The charge on capacitor is $q = CV= (2 ) \times 2 V = 8 $ μ C.

Initially
$C_1=2C$ and $C_2=C$
$q_1=2CV$
$q_2=CV$
Now condenser of capacity C is filled with dielectric K, therefore
$C'_2 = KC$
As charge is conserved
$q_1+q_2=(C'_2 + 2C)V'$
$V'=\frac {3V}{K+2}$
Hence (a) is correct

The charge on the plates are shown below


The arrangement is equivalent to four capacitors each of capacity
$C=\frac {\epsilon _0 A}{d}$
So charge in each capacitor is
$q= CV =\frac {\epsilon _0 A V}{d}$
Now for plate 1 is common to one capacitor and connected to positive terminal,so charge will be $\frac {\epsilon _0 A V}{d}$
Plate 4 is common to two capacitor and connected to negative terminal,so charge will be $\frac {-2\epsilon _0 A V}{d}$
Hence (b) is the correct option

In case of A, Potential remains same as connected to battery, Now C changes as the distance change ,so the charge changes
In case of B, charge remains same as not connected to battery, Now C changes as the distance change ,so the potential difference changes
(c) and (d)