Solutions to some interesting problems on Electrostatics of I.E. Irodov

Solutions to some interesting problems on Electrostatics of I.E. Irodov

I have often seen students seeing solutions for I.E. Irodov problems .. well its not wrong thing to look at solutions if you have tried yourself , but in all the books i have seen solutions are just given for sake of solution .. they do not focus on building concepts . And i got a chance to give some insight into concepts when i was invited by IITBANDA team to prepare an article for it .

I am adding a Disclaimer : i have taken problems from I.E. Irodov books( this is necessary as i have seen many materials copying questions from the book and do not giving due credits to them) .

One more thing If you really want to gain concept solving problems , then I.E. Irodov is must even though questions in iit are not of that level that used to come in late nineties but the concepts are same and you can only make concepts while analyzing & solving problems .

3.132

What amount of heat will be generated in the circuit shown in after the switch Sw is shifted from position I to position 2?.

Sol : Let us find first charge distribution, when switch is connected to 1

c and c_o in parallel

… (1)

Let us apply Kirchhoff’s law in the loop shown in fig.

… (2)

Solving (1) and (2)

If we carefully observe the ck + 1 and ck + 2. The only difference between two circuit is charge of polarity of the battery.

So, on change of polarity of battery, the charges on both the plate of each capacitor will just change in polarity.

Here we have to find that in charge redistribution how much charge has flown through the battery .

When switch is connected to 2

A charge q_o will flow from left most capacitor and pass through the battery and get deposited on middle capacitor.

Some amount of charge will flow from rightmost capacitor and get deposited on the middle capacitor but in this process, it does not have to pass through the battery.

Now, work done by the battery is equal to product of charge transferred and emf of the battery

Here, q_o charge has passed through the battery

So, work done by battery on the system is equal of increase in electrostatic energy stored in the circuit and heat generated.

is work done by battery on the circuit

is work done by battery on the circuit

is het generated in the circuit

Now,

Since, polarity of charges on capacitor has changed only, the magnitude of charge have remained unchanged. So, the electrostatic energy will not change

Hence,

So, heat generated

3.133

What amount of heat will be generated in the circuit shown in after the switch Sw is shifted from position 1 to position 2?

When S_w is connected to 1 charge on capacitor c is

Electrostatic energy stored on capacitor

When S_w is shifted to 2

Charge on capacitor will change to . Battery is out of the circuit and has no significance now.

Charge which has passed through the capacitor is equal to .

Work done by battery on the system

= charge transferred ×

=

Now, electrostatic energy stored in circuit

Change in electrostatic energy of circuit

=

As in last question

Heat generated

Q. 3.134  A  system Consists of two thin concentric metal shells of radii R1 and R2 with corresponding charges q1 & q2 .Find the self-energy values W1 & W2  of each shell , the interaction energy of the two shell w12 & total energy ?

Self energy of inner shell

(Refer to the article on work and energy in electrostatics)

Self energy of outer shell

Now, we have to calculate electrostatic interaction energy of the two shells.

If the electrostatic potential due to first charge configuration is same at each and every point where second charge is distributed then electrostatic interaction energy between two shells will be given by electrostatic potential due to first charge distribution at the point where second charge is distributed multiplied by total magnitude of second charge. Its vice–versa is also true

So,

where is potential due to shell on the second shell and is potential due to second shell on first shell.

Volume charge density

Electric field inside the

=

Electric field outsides the sphere

To electrostatic energy is given by

=

=

First term W₁ denote the electrostatic energy inside the charge distribution and second term W₂ denotes the electrostatic energy outside the charge distribution.

=

=

So, the electrostatic self energy of ball

Ratio of the energy w₁, stored in the ball to the energy w₂ pervading the surrounding space

Electric field inside the dielectric

Electrostatic energy stored inside the dielectric

=

=

=

3.137  A spherical Shell of Radius R1 with uniform charge q is expanded to a radius R2 . Find the work performed by the electric force in this process ?

Sol : Electrostatic energy of spherical shell of radius R,

work performed by the electric forces is at the cost of electrostatic energy

So, work done electric forces

=

=

3.138 A spherical shell of radius R ₁ with a uniform charge q hasa point charge q0 at its centre. Find the work performed by the electric forces during the shell expansion fromradius R₁ to radius R₂

W₁ = Self energy of point charge = 0

W₂ = Self energy of charge spericalshell =

W₁₂ = Electrostatic energy between point charge and charged spherical shell = potential due to point charge at the charged spherical sheel × charge on spherical shell

=

Total electrostatic energy =

=

=

Work performed by the electric forces during the shell expansion from radius to radius

process?

3.142. Inside a parallel-plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to l = 0.60 of the gap width. When the plate is absent the capacitor capacitance equals c —– 20 nF. First, the capacitor was connected in parallel to a constant voltage source producing V = 200 V, then it was disconnected from it, after which the plate was slowly removed from the gap.Find the work performed during the removal, if the plate is

(a) made of metal; (b) made of glass.

Capacitance of capacitor without plate = C (given)

Capacitance of capacitor of with plate

=

When capacitor is connected to constant voltage source producing voltage v, it will get immediately charged to c’v when voltage source is disconnected, capacitor will become isolated and its charge will remain constant when plate is removed, external agency has to do work against the electric forces as shown in fig.

When dielectric is placed two capacitor plates, it get polarized and surface charges appear on dielectric plate. These surface charges are pulled by charges on capacitor plates. So external agency pulling the dielectric plate has to apply force to pull the dielectric plate and do work on the system. The work done by external agency will be stored as electrostatic energy in capacitor. In this process, the charge on capacitor will remain constant because capacitor is isolated.

Work done by external agency in pulling the capacitor can be found out by taking difference of electrostatic energy of capacitor without plate and that of with plate

where q = c’v

This is the case when dielectric is made of glass and is dielectric constant of glass.

If dielectric is made of metal, we just have to make in our final answer.