Obj-2 Capacitors HC Verma Solutions Ch-31 Class-12 Vol-2

Obj-2 Capacitors HC Verma Solutions Ch-31 Class-12 Vol-2 Concept of Physics for Class-12. Step by Step Solutions of Objective -2 (MCQ-2) Questions of Chapter-31 Capacitors (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-12 Physics.

Obj-2 Capacitors (MCQ-2) HC Verma Solutions Ch-31 Vol-2 Concept of Physics for Class-12

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-2 (MCQ-2) Questions
Page-Number 164, 165

-: Select Topics :-

Question for Short Answer

Objective-I

Objective-II (currently open)

Exercise


Obj-2 Capacitors (MCQ-2)

HC Verma Solutions of Ch-31 Vol-2 Concept of Physics for Class-12

(Page-164)

Question-1 :-

The capacitance of a capacitor does not depend on

(a) the shape of the plates

(b) the size of the plates

(c) the charges on the plates

(d) the separation between the plates

Answer-1 :-

The option (c) the charges on the plates is correct

Explanation:

The capacitance of a capacitor is given by C=∈οA/d

Here, A is the area of the plates of the capacitor and d is the distance between the plates.
So, we can clearly see that the capacitance of a capacitor does depend on the size and shape of the plates and the separation between the plates; it does not depend on the charges on the plates.

The capacitance of a capacitor does not depend on

Question-2 :-

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

(a) The electric field in the capacitor

(b) The charge on the capacitor

(c) The potential difference between the plates

(d) The stored energy in the capacitor

Answer-2 :-

The option (b) The charge on the capacitor is correct

Explanation:

When we insert a dielectric between the plates of a capacitor, induced charges of opposite polarity appear on the face of the dielectric. They build an electric field inside the dielectric, directed opposite to the original field of the capacitor.
Thus, the net effect is a reduced electric field.
Also, as the potential is proportional to the field, the potential decreases and so does the stored energy U, which is given by U=qV/2

Thus, only the charge on the capacitor remains unchanged, as the charge is conserved in an isolated system.


Obj-2 Capacitors (MCQ-2)

HC Verma Solutions of Ch-31 Vol-2 Concept of Physics for Class-12

(Page-165)

 

Question-3 :-

A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q’.

(a) Q’ may be large than Q.

(b) Q’ must be larger than Q.

(c) Q’ must be equal to Q.

(d) Q’ must be smaller than Q.

Answer-3 :-

The option (d) Q’ must be smaller than Q. is correct

Explanation:

The relation between the induced charge Q‘ and the charge on the capacitor Q is given by Q′=Q(1-1/K)

Here, K is the dielectric constant that is always greater than or equal to 1.
So, we can see that for K > 1, Q’ will always be less than Q.

Question-4 :-

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a batter. Now,
(a) the facing surfaces of the capacitor have equal and opposite charges
(b) the two plates of the capacitor have equal and opposite charges
(c) the battery supplies equal and opposite charges to the two plates
(d) the outer surfaces of the plates have equal charges

Answer-4 :-

The options (a), (c) and (d) is correct

Question-5 :- (Obj-2 Capacitors HC Verma Solutions)

The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?
(a) Charge on the capacitor
(b) Potential difference across the capacitor
(c) Energy of the capacitor
(d) Energy density between the plates

Answer-5 :-

The options (b) and (c) are correct

Explanation:

Because the charge always remains conserved in an isolated system, it will remain the same.

Now ,

V=Qd/∈οA

Here, Q, A and d are the charge, area and distance between the plates, respectively.
Thus, as d increases, V increases.
Energy is given by

E=qV/2

So , it will also increase .

Energy density u , that is , energy stored per unit volume in the electric field is given by

u=1/2∈οE²

So , u will remain constant with increase in distance between the plates .

Question-6 :-

A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor.

(a) The battery will supply more charge.

(b) The capacitance will increase.

(c) The potential difference between the plates will increase.

(d) Equal and opposite charges will appear on the two faces of the metal plate.

Answer-6 :-

The option (d) Equal and opposite charges will appear on the two faces of the metal plate. is correct

Explanation:

Equal and opposite charges will appear on the two faces of the metal plate.

The capacitance of the capacitor in which a dielectric slab of dielectric constant K, area A and thickness t is inserted between the plates of the capacitor of area A and separated by a distance d is given by

hc verma capacitors obj 2 ans 6

Since it is given that the thickness of the sheet is negligible, the above formula reduces to hc verma capacitors obj 2 ans 6 .1 In other words, there will not be any change in the electric field, potential or charge.

Only, equal and opposite charges will appear on the two faces of the metal plate because of induction due to the presence of the charges on the plates of the capacitor.

Question-7 :- (Obj-2 Capacitors HC Verma Solutions)

Following operations can be performed on a capacitor:
X − connect the capacitor to a battery of emf ε.
Y − disconnect the battery.
Z − reconnect the battery with polarity reversed.
W − insert a dielectric slab in the capacitor.
(a) In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.
(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYZ.
(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
(d) The electric field in the capacitor after the action XW is the same as that after WX.

Answer-7 :-

The options (b), (c) and (d) are correct

Explanation:

In option (b)

If the potential is held constant, that is, the battery remains attached to the circuit, then the charge on the capacitor increases by a factor of K on inserting a dielectric of a dielectric constant K between the plates of the capacitor.

Mathematically,
q = Kq​0 
Here, q0 and q are the charges without dielectric and with dielectric, respectively.
The amount of charge stored does not depend upon the polarity of the plates.
Thus, the charge appearing on the capacitor is greater after the action XWY than after the action XYZ.

In option (c)

Since the battery is disconnected before inserting a dielectric, the amount of charge remains constant, that is,q = q​0, because after the battery is disconnected, the capacitor gets no source to store charge from. In other words, the capacitor is now an isolated system where the amount of charge is conserved and so is the energy U as 1/2q∈ . Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor. So, the energy stored in the capacitor will also not change after the action XYW.

However, during the action WXY, the amount of charge that will get stored in the capacitor will get increased by a factor of K, as the battery is disconnected after inserting a dielectric between the plates of the capacitor and the energy stored will also get multiplied by a factor of K.
Thus, the electric energy stored in the capacitor is greater after the action WXY than after the action XYW.

In option (d)

The electric field between the plates E depends on the potential across the capacitor and the distanced between the plates of the capacitor.
Mathematically,

E=∈/d

In either case, that is, during actions XW and WX, the potential remains the same, that is, ∈. Thus, the electric field E remains the same.

Denial of option (a)

During the action XYZ, the battery has to do extra work equivalent to 1/2CV² to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be 1/2CV²+1/2CV² . This extra work done will be dissipated as heat energy. Thus, thermal energy is developed. However, the stored electric energy remains unchanged, that is 1/2CV² . 

—: End of Capacitors Obj-2 (mcq-2) HC Verma Solutions Vol-2 Chapter-31 :–


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