HC Verma Questions and Solutions: Chapter 31: Capacitors- 1 | HC Verma Solutions - JEE PDF Download

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.

Here, 
Φ = Electric flux 
q = Total Charge enclosed by the Gaussian surface . 

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

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 

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 C_eq = C + C = 2C

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.

Energy density U is given by 
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

The charge induced on the plates of a capacitor is independent of the area of the plates.

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,

In this case , d = 0.
 

Region AB shows the potential difference across capacitor C​₁ 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
∴ C​​₁ < C₂ (Q remains the same in series connection.)

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.

When the spheres are connected, charges flow between them until they both acquire the same common potential V.
The final charges on the spheres are given by Q₁ = C_​1V and Q₂ = C₂V

The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows :

⇒ 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

The capacitance of a capacitor is given by 
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.

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

The relation between the induced charge Q' and the charge on the capacitor Q is given by 
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.

In H.C Verma the answer is (a), (c) ,(d). But according to us the answer should be (a), (b), (d) all these options are the properties of a capacitor and the option (c) is incorrect because the battery is a source of energy not charge. Moreover if a capacitor plates have equal charge on outside and equal charge on inside then one can think that the charge on the plates must be also equal so option (b) cant be incorrect.

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

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

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 
Since it is given that the thickness of the sheet is negligible, the above formula reduces to C =  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.

Justification of 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.
Justification of 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 (1/2)q∈. 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.
Justification of 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 (1/2)CV² to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be  This extra work done will be dissipated as heat energy. Thus, thermal energy is developed. However, the stored electric energy remains unchanged, that is