Question 1
Capacitance of a parallel plate air capacitor depends on
- Thickness of conducting plates
- Charge on the conducting plates
- Area of the conducting plates
- Distance of separation between the conducting plates
Solution
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)
Question 2
Equivalent capacitance of system of capacitors shown below in the figure is
- C/2
- 2C
- C
- None of these
Solution
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
Question 3
A parallel plate air capacitor with no dielectric between the plates is connected to the constant voltage source. How would capacitance and charge change if dielectric of dielectric constant K=2 is inserted between the plates. C
_{0 }and Q
_{0 }are the capacitance and charge of the capacitor before the introduction of the dielectric.
- C=C_{0}/2 ; Q=2Q_{0}
- C=2C_{0 }; Q=Q_{0}/2
- C=C_{0}/2 ; Q=Q_{0}/2
- C=2C_{0 }; Q=2Q_{0}
Solution
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
Question 4
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is then disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
- Zero
- 3CV^{2}/2
- 25CV^{2}/6
- 9CV^{2}/2
Solution
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
Question 5
A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field, and energy associated with this capacitor are given by Q
_{0}, V
_{0}, E
_{0 }and U
_{0} respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities are now given by Q, V, E and U are related to previous quantities as
- Q> Q_{0}
- V> V_{0}
- E> E_{0}
- U> U_{0}
Solution
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<
Question 6
The effective capacitance of a capacitor is reduced when capacitors are connected in
- series
- parallel
- series-parallel combination
- none of the above
Solution
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
Question 7
Two identical conducting plates of plate area A are separated by a distance d to form a parallel plate air capacitor. Now a metal sheet of thickness d/2 is inserted between the plates of the capacitor. The ratio of capacitance before the insertion of plate and after the insertion of plate is
- 1:2
- 1:1
- 1:2
- √2:1
Solution
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
Question 8
The plates of a parallel plate air capacitor are charged to 100 V. A 2mm plate is inserted between the plates of the capacitor. To maintain the same potential difference , the distance between the capacitor plates is increased by 1.6mm. The dielectric constant of the plate is
- 1.25
- 5
- 2.5
- 4
Solution
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
Question 9
Four capacitors are connected as shown below in the figure
Here C
_{1}=2μF , C
_{2}=3μF , C
_{3}=4μF and C
_{4}=5μF. The equivalent capacitance between A and B is
- 1.245 μF
- 4.446 μF
- 9 μF
- 3.27 μF
Solution
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
Question 10
Consider the figure and value of capacitances given in the Question 9 if a 4Volt battery is connected between points A and B total charge stored on the capacitors would be
- 13.8μC
- 10 μC
- 22 μC
- 4 μC
Solution
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
Question 11
In an isolated parallel plate capacitor of capacitance C, the four surface have charges $Q_1$, $Q_2$, $Q_3$ and $Q_4$ as shown. The potential difference between the plates is
(a) $\frac {Q_1 + Q_2 + Q_3 + Q_4}{2C}$
(b) $\frac {Q_2 + Q_3}{2C}$
(c) $\frac {Q_2 - Q_3}{2C}$
(d) $\frac {Q_1 + Q_4}{2C}$
Solution
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
Question 12
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
(a) $\frac {3CV^2}{2}$
(b) $\frac {25CV^2}{2}$
(c) $\frac {9CV^2}{2}$
(d) zero
Solution
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
Question 13
A capacitor of 4 μ F is connected as shown in the circuit . The internal resistance of the battery is 0.5 Ω . The amount of charge on the capacitor plates will be
(a) 0
(b) 4 μ C
(c) 16 μ C
(d) 8 μ C
Solution
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.
Question 14
Two condensers of capacities 2C and C are joined in parallel and charged upto potential V. The battery is removed and the condenser of capacity C is filled completely with a medium of dielectric constant K. The p.d. across the capacitors will now be
(a) $\frac {3V}{K+2}$
(b) $\frac {3V}{K}$
(c) $\frac {V}{K+2}$
(d) $\frac {V}{K}$
Solution
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
Question 15
Five identical capacitor plates each of area A are arranged such that adjacent plates are at a distance d apart . The plates are connected to a potential difference of EMF V as shown in the figure.
The charge on plates 1 and 4 will be
(a) $\frac {\epsilon _0 A V}{d}$,$\frac {2\epsilon _0 A V}{d}$
(b) $\frac {\epsilon _0 A V}{d}$,$\frac {-2\epsilon _0 A V}{d}$
(a) $\frac {-\epsilon _0 A V}{d}$,$\frac {2\epsilon _0 A V}{d}$
(a) $\frac {\epsilon _0 A V}{d}$,$\frac {-\epsilon _0 A V}{d}$
Solution
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
Question 16
A parallel plate capacitor is connected to a battery as shown below
Consider two situations:
A: Key K is kept closed and plates of capacitors are moved apart using insulating handle.
B: Key K is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
(a) In A : Q remains same but C changes.
(b) In B : V remains same but C changes.
(c) In A : V remains same and hence Q changes.
(d) In B : Q remains same and hence V changes.
Solution
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)
link to this page by copying the following text
Class 12 Maths
Class 12 Physics