Capacitors HC Verma Solutions of Que for Short Ans Ch-31 Vol-2

Capacitors HC Verma Solutions of Que for Short Ans Ch-31 Vol-2 Concept of Physics. Step by Step Solution of Questions for short answer of Ch-31 Capacitors HC Verma Question of Bharti Bhawan Publishers . Visit official Website CISCE for detail information about ISC Board Class-12 Physics.

Capacitors HC Verma Solutions of Que for Short Ans Ch-31 Vol-2

Board ISC and other board
Publications Bharti Bhawan Publishers
Ch-31 Capacitors
Class 12
Vol  2nd
writer H C Verma
Book Name Concept of Physics
Topics Solutions of Question for Short Answer
Page-Number 163

-: Select Topics :-

Que for Short Ans

Obj-1

Obj-2

Exercise


Capacitors H C Verma Que for Short Ans

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

(Page – 163)

Question 1 :-

Suppose a charge +Q1 is given to the positive plate and a charge −Q2 to the negative plate of a capacitor. What is the “charge on the capacitor”?

Answer 1 :-

Suppose a charge +Q1 is given to the positive plate and a charge −Q2 to the negative plate of a capacitor. What is the "charge on the capacitor"?

Given :

Charge on the positive plane = +Q1

Charge on the negative plate = -Q2

To calculate: Charge on the capacitor

Let ABCD be the Gaussian surface such that faces AD and BC lie inside plates X and Y, respectively.
Let q be the charge appearing on surface II. Then, the distribution of the charges on faces I, III and IV will be in accordance with the figure.

Let the area of the plates be A and the permittivity of the free space be ∈0.

Now, to determine q​ in terms of Q1 and Q2, we need to apply Gauss’s law to calculate the electric field due to all four faces of the capacitor at point P. Also, we know that the electric field inside a capacitor is zero.

Electric field due to face I at point P, E1  hc verma capcitors short ans img 1

Electric field due to face II at point P, E2= hc verma capcitors short ans img 2

Electric field due to face III at point P, E3 = hc verma capcitors short ans img 3

Electric field due to face IV at point P, E4 =hc verma capcitors short ans img 4    (Negative sign is used as point P lies on the LHS of face IV.)

Since point P lies inside the conductor,

E1 + E2 + E3 + E4 = 0

∴ Q1 – q + q – q – ( -Q2 + q) = 0

⇒ q = hc verma capcitors short ans img 5

Thus , the change on the capacitor is hc verma capcitors short ans img 5, which is the charge on faces  II and III.

Question 2 :-

As C=(1/V)Q , can you say that the capacitance C is proportional to the charge Q?

Answer 2 :-

No. Since capacitance is a proportionality constant, it depends neither on the charge on the plates nor on the potential. It only depends upon the size and shape of the capacitor and on the dielectric used between the plates.
The formula that shows its dependence on the size and shape of the capacitor is as follows : 

C=∈οA/d 

Here, A is the area of the plates of the capacitor and d is the distance between the plates of the capacitor.

Question 3 :- (Capacitors HC Verma Solutions)

A hollow metal sphere and a solid metal sphere of equal radii are given equal charges. Which of the two will have higher potential?

Answer 3 :-

The potential of a metal sphere is directly proportional to the charge q given to it and inversely proportional to its radius r. i.e. V = q/4π∈οr

Since both the spheres are conductors with the same radius and charge, the charge given to them appears on the surface evenly. Thus, the potential on the surface or within the sphere will be the same, no matter the sphere is hollow or solid.

Question 4 :-

The plates of a parallel-plate  capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces?

Answer 4 :-

It is given that the plates of the capacitor have the same charges. In other words, they are at the same potential, so the potential difference between them is zero.

The plates of a parallel-plate  capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces?

Let us consider that the charge on face II is so that the induced charge on face III is -q and the distribution is according to the figure.

Now, if we consider Gaussian surface ABCD, whose faces lie inside the two plates, and calculate the field at point P due to all four surfaces, it will be

hc verma capcitors short ans img 6

(It is –Ve because point P is on the left side of face IV. )

Now, as point P lies inside the conductor, the total field must be zero.

∴ E1 + E2 + E3 + E4 = 0

Or

Q – q + q – q + Q+ q = 0

∴ q = 0

Hence, on faces II and III, the charge is equal to zero; and on faces I and IV, the charge is Q.
Thus, it seems that the whole charge given is moved to the outer surfaces, with zero charge on the facing surfaces.

Question 5 :-

A capacitor has capacitance C. Is this information sufficient to know what maximum charge the capacitor can contain? If yes, what is this charges? If no, what other information is needed?

Answer 5 :-

No. This information is not sufficient. Since the charge is proportional to the potential difference across the capacitor, we need to know the potential difference applied across the capacitor.

q ∝ V ⇒ q = CV

Here, q is the charge, V is the potential difference applied and C is the proportionality constant, i.e. capacitance.

 

Question 6 :- (Capacitors HC Verma Solutions)

The dielectric constant decreases if the temperature is increased. Explain this in terms of polarization of the material.

Answer 6 :-
The amount of polarisation can be understood as the extent of perfect alignment of the molecules of a dielectric with an external electric field. The more aligned the molecules are with the external magnetic field, the more is the polarisation and the more will be the dielectric constant.
But with increase in temperature, the thermal agitation of the molecules or the randomness in their alignment with the field increases.
Thus, we can say that increase in temperature results in reduced polarisation and reduced dielectric constant.

Question 7 :-

When a dielectric slab is gradually inserted between the plates of an isolated parallel-plate capacitor, the energy of the system decreases. What can you conclude about the force on the slab exerted by the electric field?

Answer 7 :-

As the energy of the system decreases, the change in the energy is negative. Force is defined as a negative rate of change of energy with respect to distance.

hc verma capcitors short ans img 7

So, as the energy decreases, the force due to the electric field of the capacitor increases when the dielectric is dragged into the capacitor.

You might also like
Leave a comment