HC Verma Physics Part 2 Solutions for Chapter 31 Capacitor in PDF

HC Verma Physics Part 2 Solutions for Chapter 31 Capacitor

5/21/2012

Download HC Verma Solutions for HC Verma Physics Part 2 Solutions for Chapter 31 Capacitor solved by our expert teachers. We have curated solutions for all questions of chapter 31. You can download the Solution of HC Verma - Capacitor and prepare for your upcoming competitive examinations.

Solution of HC Verma Concept of Physics -2- Chapter 31 Capacitor

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221 Comments

Asp

6/13/2012 05:03:57 pm

I have some problem in capacitor

Admin

6/13/2012 05:37:06 pm

Please ask we are here to help you.

ayesha link

6/20/2012 04:03:36 am

i had prob in qurs 25 (a) of chapter 31 i.e. capacitors.......didnt understand the solution given here

arpit jain

6/28/2015 02:23:25 am

i have no problem

Vatsal

7/14/2016 10:25:53 pm

I have problem in question 57

medha

7/3/2018 08:11:46 am

Question 21 hc verma page 166

anupama

6/13/2013 04:18:12 pm

i didn'tundestand solution of que25(a) in capacitor

mruganshu

6/22/2014 02:26:56 pm

i also have same prob so if u get then please give it to me also

Apurv Tiwari

1/11/2015 09:37:47 pm

I prefer to combine all the capacitors. Use the series and parallel formulas to get equivalent capacitance. I can't draw the figure here so I have named all the capacitors as follows.

I am naming the capacitors. Let the Cab one of 4 UF be called C1. The Ca capacitor be called C2. The CD capacitor be called C3 and the last BC capacitor be called C4.

Then combine them. C1, C2 series gives capacitor C5 = 4/3 UF
Then C5 and C3 combine in parallel to give C6 = 16/3 UF.
Then C6 and C4 combine to give net capacitance C7 = 16/11 UF. Thus, Q7 = 12X(C7) = 192/11 UC.

Then do the reverse. As C6 and C4 were in series, they share same charge = Q7.
Then break C6 into C3 and C5. They are in parallel, so they share charge in the ratio of their capacitances. Let C5 have charge Q5. Thus, C3 has (Q7 - Q5) charge.
Thus, [(Q7 - Q5)/C3] = [Q5/C5]
Solving this gives Q5 = 48/11 UC.

Again, break C5 into C1 and C2. Being in series, they share the same charge Q5.

So, Va - Vb = Q5/C1 by the basic formula of capacitors.
Solving this gives Va - Vb = 12/11 V.

Try to follow each step in your notebook and you will surely understand what i am trying to say.

NARESH KNAOJIA

4/9/2016 08:20:56 am

kirchoff's law is applied as we do in resistance
the question is then solved according to required subsitution

subhransu S bhatt

6/14/2016 12:55:34 pm

the question as well as the answer is wrong the problem of the book is different..... the answer is 15uV(i have got the answer checked by verma sir himself)..

anupama

6/13/2013 04:20:45 pm

i didn't undestand solution of que25(a) in capacitor

shushant

12/9/2014 02:26:03 pm

it is basicallly one by applying kvl evn idk wht they did in the solution bt by applying kvl i got the answr

jazz link

10/18/2015 02:13:53 am

mais je ne comprends pas(q-25(b)) .....Je ne crois pas qu'ils comprenaient cela,

shreya

5/27/2016 05:00:39 am

cant understand ques 23 n 22 of capacitance

anu

7/11/2016 06:41:38 am

arjun

6/1/2018 05:35:51 pm

I have problem in question 25(b)

medha link

7/3/2018 08:09:30 am

I have problem in question21 ,page 166 hc verma

Ab link

8/2/2018 01:22:24 am

i hate u

ayesha link

6/20/2012 04:05:03 am

i had prob in qurs 25 (a) of chapter 31 i.e. capacitors.......didnt understand the solution given here

Admin

6/21/2012 11:32:11 am

In question no. 25(a) they have simple applied kirchhoff's voltage law (KVL) please refer your textbook to know more about KVL.
- Be in touch we are always here..

ayesha link

6/22/2012 12:44:13 am

yaa dats ok.................bt i guess d solution is some what wrong

supriya

8/27/2014 08:17:38 pm

ayesha link

6/22/2012 12:46:34 am

yaa dats ok.................bt i guess d solution is some what wrong

ayesha link

6/22/2012 12:48:43 am

yaa dats ok.................bt i guess d solution is some what wrong!!!!!

ayesha link

6/22/2012 12:49:26 am

i dont noe in d second loop fr second equation i didnt get dat

Admin

6/22/2012 01:10:32 pm

yaa you are correct, I too find problem in second loop ...
simply solve it using KVL..

anonymous

3/30/2013 02:53:06 pm

Q.25 (a) draw the equivalent circuit of the figure.(it is very easy method and time saving as well)

Admin

11/3/2012 03:49:37 pm

This is attempt to provide free education to children who are preparing for engineering entrance examination.
If you are having any doubt regarding solving questions of HC Verma or understanding any topic related to this, you may leave your comment.
we will give your answer within 24 hours.
Thank You

Prachi

3/21/2013 10:05:13 pm

in question 66 how you derived the force on dielectric slab?

theyatin

3/21/2013 11:48:30 pm

that's the formula of force exerted on the dielctric slab inserted between capacitor.

anonymous

3/30/2013 02:57:07 pm

can anyone please explain star delta transformation used in Q.26 (a)

theyatin

3/30/2013 03:35:23 pm

dear,
lets consider the lower part of the circuit as

c1 __ c2
c3 c4

then the delta function is as
(c1+c3) (c2+ c4)/ [(c1+c3)+(c2+c4)]
+ c (upper part of circuit)

Apurv Tiwari

1/11/2015 09:39:02 pm

Even I didn't understand that!

suyash

4/16/2013 04:44:04 am

Q NO 30 mein net charge kya hai?

theyatin

4/17/2013 08:43:35 pm

dear,
the NET charge is Mean of difference of charge of each plate
as here Q2 is negative so Q1+Q2 /2

suyash

4/21/2013 04:06:19 am

why we take mean of difference

theyatin

4/21/2013 01:32:21 pm

dear,
to find mean charge at the center of capacitor due to both plates.

sai

4/25/2013 02:54:48 am

i want solution for 39 problem

theyatin

4/27/2013 01:59:34 pm

Dear Sai,
Please note in the question that identical capacitor is connected. so, as the capacitor is identical and energy is constant. so, energy will be divided equally.

manvi

5/10/2013 03:04:40 am

Two insulated charged sphere of radii R1andR2having charges Q1 and Q2respectively are connected to each other then there is
1) NO CHANGE IN THE ENERGY SYSTEM
2 a decrease in the energy system unless q1R
2 =q2R1

paras

5/22/2013 09:37:25 pm

Please explain the role of the switch in q 46, when open and closed ??

theyatin

5/22/2013 10:21:38 pm

dear,
when switch is opened the capacitors are in series.
when close capacitors are in parallel.

aniruddh

5/25/2013 01:55:54 am

cannot understand how the plates in series for hcv part2 chapter-31,q.no33

theyatin

5/26/2013 12:56:15 am

dear,
rotate book at 90'
now rotate each plates to 90' attach wire to the middle of the plate rather than at the lower end of plate you can see how these plates are in series.

deepa link

10/17/2015 09:38:14 pm

sorry cant understand!..........

arghadeep

5/29/2013 10:24:53 pm

i cant understand q 57 any1 please help

theyatin

5/30/2013 01:26:02 am

dear,
in the expression given in solution doesnt have any measure of position thus position of metal doesnt matters.
apart from this equation.
thickness of metal as well as distance between position matters but it doesnt matters if the metal is closer to A plate or B plate.
thus position dosnt matters.

arghadeep

6/2/2013 02:42:34 am

ok thank you i got it.i need help with q 61 and 68

Prateek

5/30/2013 05:14:11 am

Sir please tell me Objective exercise question 7..
and question for short ans Ques 3..In this question i cant understand which potential they have asked us to compare..the potential when distance < R or > R..

theyatin

5/30/2013 05:50:31 pm

dear,
which objective dear 1 or 2??
in short answer 3
both of the bodies will have same potential out of the body as well as on the body.
but there is no potential inside hollow spherical shell but there is some potential inside solid sphere.

anonymous

5/30/2013 04:22:25 pm

Ques 45 AND 47

How to calculate charges after reconnection?

Also, why is the formula C1V1 - C2V2 / C1+C2 for common potential V and not C1V1 + C2V2/C1+C2?

theyatin

5/30/2013 06:24:21 pm

dear,
as we need to calculate potential difference we have to get effective potential by subtracting them.

Mohit

6/20/2018 10:43:13 am

I still don't get it C1*V1 will give charge which needs to add up not subtracted

ub.

6/1/2013 05:32:05 pm

I have problem in ...chapter 31 'capacitor'...question 53 (c) part it says that "dielectric is inserted into the gap to occupy lower half of it" what does this mean in solution u have not considered half plate in capacitor

theyatin

6/2/2013 12:26:31 am

dear,
as dielectric slab is touching lower plate as the charge is transferred on top of dielectric plate too so the distance between both charges is half as its distance from other plate.

6/3/2013 12:20:46 am

sry sir ....I didn't understant

atul

6/2/2013 01:42:35 am

i can"t understand ques no 25 b

theyatin

6/2/2013 09:12:35 pm

dear,
as both batteries are connected in opposite directio n to each other so potenial difference is 12 V
now the flow of current would be only from 12 V to @4V through 4 farade capacitor.
thus 0*2+12*4/2+4 (as there is no potential difference anticlockwise)

arghadeep

6/2/2013 02:36:55 am

som1 plase help me with q 61 and 68 .please email me to [email protected]

arghadeep

6/2/2013 02:37:07 am

som1 plase help me with q 61 and 68 .please email me to [email protected] .

arghadeep

6/2/2013 02:38:22 am

som1 plase help me with q 61 and 68 .

theyatin

6/2/2013 09:47:10 pm

dear,
in 61
first we calculate capacitance between a and c
then we calculate between B to c
now as both capacitors are in parallel
thus capacitance is 1/C=1/Cac + 1/Cbc
as capacitance of a to c is due to dielectric material . . thus capacitance is multiplied by K that is dielectric constant.

6/3/2013 12:25:29 am

dear sir .....in question 55 (a) part I have calculated C(without dielectric) and C' (with dielectric) and then calculated U1=1/2CV^2 U2=1/2C'V2 then U2-U1 but it is not giving increase in electrostatic Energy.......and u have not solved it.....please help

theyatin

6/4/2013 01:25:51 am

dear,
we regret this mistake. our technical team will soon get this problem solved.
however, in solution answer of a) is given but not calculated it is 7.08uF = 7.1uF
b).
when the dielectric slab is taken out so t and t/k in denominator are =0
thus answer can be seen as in solution C= 3.54uF

theyatin

6/4/2013 01:29:29 am

dear,
in Q53 c)
the term d-t = 10^-3
as d=2* 10^-3 and t=10^-3
so d-t= 10^3

archit

6/9/2013 04:55:27 pm

plzz give some other solution for q25(b) ......... i cant understand the solution given above .... plz help

Admin

6/13/2013 10:42:12 pm

Yatin sir is not well. Please wait 1-2 days.

theyatin

6/15/2013 04:23:09 pm

dear,
as both batteries are connected in opposite direction n to each other so potential difference is 12 V (24-12)
now the flow of current would be only from 12 V to @4V through 4 farad capacitor.
thus 0*2+12*4/2+4 (as there is no potential difference anticlockwise)

Pal

6/9/2013 08:46:41 pm

What is the equivalent capacitance of two spheres of radius a,b separated by a distance d d>> a,b.

Admin

6/13/2013 10:55:10 pm

Yatin sir is unwell. Please wait 1-2 days.

theyatin

6/15/2013 05:48:22 pm

dear.
capacitance becomes negligible over large distances.
so there would be negligible capacitance

Abhi

6/12/2013 07:23:03 pm

plz help me for q. 23
also i don,t understand , in q 18 , in solution , what is in[4/2] ?

admin

6/13/2013 10:56:15 pm

Yatin sir is not well. Please wait 2 days.

theyatin

6/15/2013 05:20:20 pm

dear,
as F=ma=qE
so acceleration of electron = qE/m where m is mass of electron
no from S=ut+1/2at^2
X=1/2at^2 as u =0
so now dividing distance traveled by proton and electron =ap/ac
so now we can put values of acceleration of proton and acceleration of electron, now you can solve this to get answer X

theyatin

6/15/2013 05:46:24 pm

dear in Q18
there is no [4/2] in question please ask your question again . . .

Keshav

6/12/2013 09:18:32 pm

Please explain how we derived v=w/q in 7ques b part in detail please

admin

6/13/2013 10:58:39 pm

Yatin sir is not feeling well. Please wait 1-2 days.

Keshav

6/15/2013 02:57:21 am

Also explain why we are deviding 8.52 by 4 in ques50 c part

theyatin

6/15/2013 05:28:32 pm

dear.
as there are four equally charged surfaces of box thus to find charge over one surface we divide it by 4.

Keshav

6/16/2013 01:05:17 am