Capacitors Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-31
Capacitors Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-31 Concept of Physics for Class-12. Step by Step Solutions of Objective -1 (MCQ-1) Questions of Chapter-31 Capacitors (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Capacitors Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-31
| Board | ISC and other board |
| Publications | Bharti Bhawan Publishers |
| Chapter-31 | Capacitors |
| Class | 12 |
| Vol | 2nd |
| writer | HC Verma |
| Book Name | Concept of Physics |
| Topics | Solution of Objective-1 (MCQ-1) Questions |
| Page-Number | 164 |
-: Select Topics :-
Obj-1
Obj-2
Exercise
Capacitors Obj-1 (MCQ-1) Questions
HC Verma Solutions of Ch-31 Vol-2 Concept of Physics for Class-12
(Page-164)
Question-1 :-
A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is
(a) CV/∈ο
(b) 2CV/∈ο
(c) CV/2∈ο
(d) Zero
Answer-1 :-
Since the net charge enclosed by the Gaussian surface is zero, the total flux of the electric field through the closed Gaussian surface enclosing the capacitor is zero.
Question-2 :-
Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be
(a) 2 C and 2 V
(b) C/2 and V/2
(c) 2 C and V/2
(d) C/2 and 2 V.
Answer-2 :-
Since the voltage gets added up when the capacitors are connected in series, the voltage of the combination is 2V.
Also, the capacitance of a series combination is given by
Here ,
C_net = Net capacitance of the combination
C1 = C2 = C
∴ Cnet = C/2
Question-3 :-
If the capacitors in the previous question are joined in parallel, the capacitance and the breakdown voltage of the combination will be
(a) 2 C and 2 V
(b) C and 2 V
(c) 2 C and V
(d) C and V.
Answer-3 :-
2C and V
In a parallel combination of capacitors, the potential difference across the capacitors remain the same, as the right-hand-side plates and the left-hand-side plates of both the capacitors are connected to the same terminals of the battery. Therefore, the potential remains the same, that is, V.
For the parallel combination of capacitors, the capacitance is given by
Ceq = C1 +C2
Here ,
C1 = C2 = C
∴ Ceq = 2C
Question-4 :-
The equivalent capacitance of the combination shown in the figure is .
(a) C
(b) 2 C
(c) C/2
(d) none of these.
Answer-4 :-
Since the potential at point A is equal to the potential at point B, no current will flow along arm AB. Hence, the capacitor on arm AB will not contribute to the circuit. Also, because the remaining two capacitors are connected in parallel, the net capacitance of the circuit is given by
Ceq = C + C = 2C
Question-5 :- (Capacitors Obj-1 HC Verma)
A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will
(a) increase
(b) decrease
(c) remain unchanged
(d) become zero
Answer-5 :-
The force between the plates is given by 
Since the capacitor is isolated, the charge on the plates remains constant. We know that the charge is conserved in an isolated system. Thus, the force acting between the plates remains unchanged.
Question-6 :-
The energy density in the electric field created by a point charge falls off with the distance from the point charge as
(a) 1/r
(b) 1/r²
(c) 1/r³
(d) 1/r4
Answer-6 :-
The electric field created by a point charge at a distance r is given by 
On putting the above form of E in eq. 1, we get
Thus, U is directly proportional to 1/r4
Question-7 :-
A parallel-plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let Q, and Q be the charges appearing on the positive and negative plates respectively.
(a) Q+ > Q –
(b) Q+ = Q –
(c) Q+ < Q –
(d) The information is not sufficient is decide the relation between Q, and Q.
Answer-7 :-
The charge induced on the plates of a capacitor is independent of the area of the plates.
Question-8 :-
A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates . The capacitance now becomes.
(a) C/2
(b) 2 C
(c) 0
(d) ∞
Answer-8 :-
The thin metal plate inserted between the plates of a parallel-plate capacitor of capacitance C connects the two plates of the capacitor; hence, the distance d between the plates of the capacitor reduces to zero. It can be observed that the charges on the plates begin to overlap each other via the metallic plate and hence begin to conduct continuously.
Mathematically ,
Question-9 :- (Capacitors Obj-1 HC Verma)
The following figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors.
(a) C1 > C2
(b) C1 = C2
(c) C1 < C2
(d) The information is not sufficient to decide the relation between C1 and C2.
Answer-9 :-
Region AB shows the potential difference across capacitor C1 and region CD shows the potential difference across capacitor C2. Now, we can see from the graph that region AB is greater than region CD. Therefore, the potential difference across capacitor C1 is greater than that across capacitor C2.
∵ Capacitance, C = Q/V
∴ C1 < C2 (Q remains the same in series connection.)
Question-10 :-
Two metal plates having charges Q, −Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will
(a) increase
(b) decrease
(c) remain the same
(d) become zero
Answer-10 :-
Oil between the plates of the capacitor acts as a dielectric. We know that the electric field decreases by a factor of 1/K of the original field when we insert a dielectric between the plates of a capacitor (K is the dielectric constant of the dielectric). So, if the oil is pumped out, the electric field between the plates will increase, as the dielectric has been removed.
Question-11 :-
Two metal spheres of capacitance C1 and C2 carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy.
(a) 
(c) 
(d) 
Answer-11 :-
is correctThe final charges on the spheres are given by
Q1 = C1V and Q2 = C2V
∴ Q1/Q2 = C1V / C2V=C1/C2
Question-12 :- (Capacitors Obj-1 HC Verma)
Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are
(a) 6 µF, 18 µF
(b) 3 µF, 12 µF
(c) 2 µF, 12 µF
(d) 2 µF, 18 µF
Answer-11 :-
The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows :
1/C = 1/6+1/6+1/6 = 1/2
⇒ C =2 µF
The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows :
C = 6+6+6 = 18 µF
—: End of Capacitors Obj-1 (MCQ-1) HC Verma Solutions Vol-2 Chapter-31 :–
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