For parallel plate air capacitor
C = (ε_{0}A)/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 C_{0 }and Q_{0 }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=Kε_{0}
Where K is dielectric constant of the material.
Thus,
C = (Kε_{0}A)/d
= K C_{0}
Or, C= 2 C_{0}
Again charge on the capacitor after the introduction of dielectric is given as
Q=CV (a)
And before the introduction of dielectric
Q_{0}= C_{0}V
Now putting the value of C in equation (a) we find
Q=2 C_{0}V
Or, Q=2 Q_{0}
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 Q_{1}=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 = 3CV^{2}/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 V_{0}. Introduction of dielectric increases the capacitance C and hence charge stored in the capacitor also increases since Q=CV thus Q>Q_{0}. 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=CV^{2}/2. Thus U>U_{0}.
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 = (ε_{0}A)/d
And after the insertion of any dielectric plate filling the capacitor partially is
C = (ε_{0}A)/[(d-t)+t/K]
We know that K for a metal is infinite thus
C = (ε_{0}A)/(d-t)
Here t=d/2 thus
C’=(2ε_{0}A)/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 C_{1 }and series combination of C_{2} , C_{3} and C_{4} 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=C_{1}+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)