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Chapterwise Solutions of HC Verma

HC Verma Physics Part 2 Solutions for Chapter 31 Capacitor

5/21/2012

221 Comments

 
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

Download HC Verma Solutions for Chapter 31 for Free in PDF

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221 Comments
Asp
6/13/2012 05:03:57 pm

I have some problem in capacitor

Reply
Admin
6/13/2012 05:37:06 pm

Please ask we are here to help you.

Reply
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

Reply
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

Reply
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

Reply
anu
7/11/2016 06:41:38 am

Reply
arjun
6/1/2018 05:35:51 pm

I have problem in question 25(b)

Reply
medha link
7/3/2018 08:09:30 am

I have problem in question21 ,page 166 hc verma

Reply
Ab link
8/2/2018 01:22:24 am

i hate u

Reply
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

Reply
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..

Reply
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

Reply
ayesha link
6/22/2012 12:48:43 am

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

Reply
ayesha link
6/22/2012 12:49:26 am

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

Reply
Admin
6/22/2012 01:10:32 pm

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

Reply
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)

Reply
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

Reply
Prachi
3/21/2013 10:05:13 pm

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

Reply
theyatin
3/21/2013 11:48:30 pm

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

Reply
anonymous
3/30/2013 02:57:07 pm

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

Reply
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)

Reply
Apurv Tiwari
1/11/2015 09:39:02 pm

Even I didn't understand that!

Reply
suyash
4/16/2013 04:44:04 am

Q NO 30 mein net charge kya hai?

Reply
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

Reply
suyash
4/21/2013 04:06:19 am

why we take mean of difference

Reply
theyatin
4/21/2013 01:32:21 pm

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

Reply
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

Reply
paras
5/22/2013 09:37:25 pm

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

Reply
theyatin
5/22/2013 10:21:38 pm

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

Reply
aniruddh
5/25/2013 01:55:54 am

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

Reply
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.

Reply
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

Reply
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.

Reply
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..

Reply
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.

Reply
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?

Reply
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.

Reply
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

Reply
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.

Reply
ub
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

Reply
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)

Reply
arghadeep
6/2/2013 02:36:55 am

som1 plase help me with q 61 and 68 .please email me to arghadeep2496@gmail.com

Reply
arghadeep
6/2/2013 02:37:07 am

som1 plase help me with q 61 and 68 .please email me to arghadeep2496@gmail.com .

Reply
arghadeep
6/2/2013 02:38:22 am

som1 plase help me with q 61 and 68 .

Reply
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.

Reply
ub
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

Reply
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

Reply
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

Reply
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

Reply
Admin
6/13/2013 10:42:12 pm

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

Reply
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)

Reply
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.

Reply
Admin
6/13/2013 10:55:10 pm

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

Reply
theyatin
6/15/2013 05:48:22 pm

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

Reply
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] ?

Reply
admin
6/13/2013 10:56:15 pm

Yatin sir is not well. Please wait 2 days.

Reply
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

Reply
theyatin
6/15/2013 05:46:24 pm

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

Reply
Keshav
6/12/2013 09:18:32 pm

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

Reply
admin
6/13/2013 10:58:39 pm

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

Reply
Keshav
6/15/2013 02:57:21 am

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

Reply
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

in ques50 c part there are 6 sides so we need to devide the total charge by six (coated surface also include ) why we are taking 4 sides
Also please explain Q 46 what is the role of switch

Keshav
6/16/2013 01:09:03 am

in ques50 c part there are 6 sides so we need to devide the total charge by six (coated surface also include ) why we are taking 4 sides
Also please explain Q 46 what is the role of switch

theyatin
6/15/2013 05:24:49 pm

dear,
work done in taking a charge q over voltage potential V is given as
W=V*q

Reply
Keshav
6/16/2013 01:03:35 am

in ques50 c part there are 6 sides so we need to devide the total charge by six (coated surface also include ) why we are taking 4 sides
Also please explain Q 46 what is the role of switch

Reply
theyatin
6/16/2013 11:28:58 pm

dear,
dear there are only 4 sides top and bttom is empty.

Reply
Keshav
6/16/2013 01:03:43 am

in ques50 c part there are 6 sides so we need to devide the total charge by six (coated surface also include ) why we are taking 4 sides
Also please explain Q 46 what is the role of switch

Reply
theyatin
6/16/2013 11:30:48 pm

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

Reply
Keshav
6/16/2013 01:03:57 am

in ques50 c part there are 6 sides so we need to devide the total charge by six (coated surface also include ) why we are taking 4 sides
Also please explain Q 46 what is the role of switch

Reply
Keshav
6/16/2013 01:05:30 am

in ques50 c part there are 6 sides so we need to devide the total charge by six (coated surface also include ) why we are taking 4 sides
Also please explain Q 46 what is the role of switch

Reply
Keshav
6/16/2013 01:09:13 am

in ques50 c part there are 6 sides so we need to devide the total charge by six (coated surface also include ) why we are taking 4 sides
Also please explain Q 46 what is the role of switch

Reply
sayan
6/16/2013 02:46:46 am

Sir, please explain Q 25(b). I can't understand how you find the P.D.

Reply
theyatin
6/16/2013 11:17:42 pm

dear sayan,
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)

Reply
Keshav
6/16/2013 02:53:47 am

Sir in Q 40 why have we taken v as constant when the two capacitors are joined

Reply
theyatin
6/16/2013 11:26:10 pm

dear keshav,
V=Q1/C1
V=C2/V2
hence both are equal and can be compared.

Reply
Keshav
6/16/2013 03:33:46 am

Sir in Q 40 why have we taken v as constant when the two capacitors are joined also the ans is wrong of Q 42

Reply
shivam
6/23/2013 12:54:38 am

Pls bother to tell me , what is dielectric is made of and its effect on inserting it in parallel plate capacitor ,thanks

Reply
theyatin
6/23/2013 10:42:29 pm

dear,
A dielectric is an electrical insulator that can be polarized by an applied electric field
The most obvious advantage to using such a dielectric material is that it prevents the conducting plates on which the charges are stored from coming into direct electrical contact.
Mica is an example of solid dielectric material or some kind of oils are also used as liquid dielectric materials.

Reply
ab
6/23/2013 07:23:14 pm

in ques no 45, why is ther some charge flowing from the battery, although the capacitor is fully charged and is also at a higher potential in comparison to the 12V battery?

Reply
theyatin
6/23/2013 10:43:22 pm

dear,
which part of Q 45 you are asking for??

Reply
ab
6/24/2013 02:47:46 am

dear,
b part

Srishti
6/29/2013 04:31:03 pm

could u please expain me ques no 66?

Reply
srishti
6/29/2013 04:32:45 pm

could u please explain me ques no 66

Reply
srishti
6/29/2013 04:34:48 pm

could u please explain ques no 66

Reply
thyatin
7/1/2013 01:35:44 am

dear,
as dielectric is attracted by electric field.
which is balanced by weight of block.
so putting force due to electric field equal to Mg we can have mass.

Reply
Anand
7/2/2013 08:10:09 am

Please explain Q no. 58. I think the solution provided here for Q. 58 is wrong. Please explain what should be the correct procedure.

Reply
theyatin
7/3/2013 04:05:54 pm

dear,
what are you asking for. . answer is wrong or solution is wrong??

Reply
Anand
7/4/2013 12:44:04 pm

I have understood the solution. Its okay, I don't need help on that question anymore. Thank you.

anupam
7/20/2013 12:37:43 am

what is the formula of star delta conversion in capacitor

Reply
Vallabh Neema
7/30/2013 04:05:50 pm

q.no.8,33,40(d),43,50

Reply
theyatin
7/31/2013 05:03:06 am

dear,
B and C are parallel so they will add up 4+4
and then A and B+C are in siries so they will add harmonically
such as A*(B+C) / A+(B+C)
that is 4.
so now q=CV
as voltage is divide half so it is 6 volt.as voltage is divided in series.

Reply
theyatin
7/31/2013 05:05:01 am

dear,
what problem do you have in Q33?
is it how they are in series?

Reply
theyatin
7/31/2013 05:06:14 am

dear,
40th Q dont have parts.

Reply
priya link
8/5/2013 08:16:35 pm

i want solution of question 33 chapter capacitors of hc verma , can u pls help??

Reply
theyatin
8/7/2013 10:19:13 pm

dear,
which Question you want to be answered??

Reply
shivam
8/9/2013 02:27:34 pm

how do we know in question 17
whether capacitors are connected in series or parallel ?

Reply
theyatin
8/12/2013 01:28:28 am

dear,
capacitors are parrallel. . .

Reply
rashi
8/9/2013 02:31:49 pm

yes , i wanted to know the same

Reply
shabd kumar
9/21/2013 02:23:39 pm

sir, please tell me que. 9 of objective 1, que.20 of exercise.

Reply
shabd kumar
9/22/2013 07:02:45 am

sir, plz explain me que. 9 of objective1 and que 20 of exercise.

Reply
theyatin
9/23/2013 02:57:24 am

dear,
as the particle completes whole circle thus work done is force applied * displacement
that is 0
so work done by electric field is zero.

Reply
shabd kumar
9/22/2013 12:50:34 pm

plz give the solution of que.22 of exercise.

Reply
shabd kumar
9/23/2013 03:00:43 pm

sir, please give the solution of max 3 que. in one day as rule.
Plz show the equivalent capacitance in que.33 of exercise

Reply
theyatin
9/25/2013 04:02:13 am

dear,
C=e A/d
as
solving C by putting values
we have capacitance for on capacitor
C=2.4*10^-9F
thus as all these capacitors are in series
thus C=C/3
hence you can find capacitance for single capacitor

Reply
shabd kumar
9/24/2013 03:39:04 pm

plz, give the solution of que.22 of exercise

Reply
theyatin
9/29/2013 03:47:47 am

dear,
first we need to use formula of verticle distance covered by particle that is d1/2
now we can find E=V/R
also for capcitor A
V1
now total C from d1+d2
hence finding q
and putting its value in E
substituting value of E in verticla distance covered by projectile in first equation
we can find u^2

Reply
shabd kumar
9/26/2013 06:47:02 am

plz, give the solution oe que.22 of exercise

Reply
Shabd
9/28/2013 12:02:10 pm

Plz, give the solution of que.57 of exercise

Reply
shabd
9/28/2013 04:26:39 pm

plz, give the solution of que.58 of exercise.

Reply
theyatin
9/29/2013 03:54:10 am

Q58
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.

Reply
shabd
9/28/2013 04:34:55 pm

plz, give the solution of que.67 of exercise.

Reply
shabd
9/28/2013 04:58:49 pm

in que68 of exe. How will we find angular velocity of slab to estimate its time period. Plz, explain me.

Reply
theyatin
9/29/2013 04:08:57 am

dear,
we have found the time you dont need to calculate angular velocity

Reply
shabd
9/29/2013 06:56:41 pm

Is the solution of que58 that u posted correct?
Explain me sir

Reply
theyatin
10/1/2013 03:24:17 am

dear,
pardon me i have mistakenly explained wrong question
a)
when switch is closed the energy may correspond to both capacitors as they are parallel
b)
when switch is opened then capacitance will be divided between two capacitors
now E=1/2 (cv)^2 /Ceff
ceff=3c
hence
calculating capacitance for other capcitor E=1/2 *Ceff v^2
hence adding up we can have final energy
so now intial energy/. /final energy =3

Reply
deep
10/20/2013 01:57:56 pm

sir, can we conclude that every time a charged capacitor is disconnected and then connected to a battery with reversed polarity, then the work done by battery will be positive and negative if the polarity remains same???

Reply
theyatin
10/31/2013 02:55:02 am

dear,
work done can not be both positive and negative at same time

Reply
Ani
10/30/2013 05:19:17 pm

A capacitor of capacitance c is charged to a potential difference v from a cell and then disconnected from
it. A charge +Q is given to its positive plate. the potential difference across the capacitor is now:-
(a)V
(b)v+QlC
(c)v+Q/2C
(d)v-Q/C'ifQ<Cv'
sir can u please explain me this one..!! :)

Reply
theyatin
10/31/2013 03:00:55 am

dear,
if positive side is added up with charge +Q
the potential would increase

so the added up potential would be V=Q/C
thus answer may be v+Q/C

Reply
ani
11/1/2013 11:44:26 am

thankyou sir..!! :)

Anonymous
11/4/2013 04:13:29 am

Sir,
Please can you explain question 8 of exercise, as in why have they taken voltage as 6.

Reply
theyatin
11/4/2013 11:40:59 pm

dear,
voltage is divided half between q1 and q2 thus equal on q2 and q3.

Reply
Saswato Sen
11/25/2013 11:29:28 pm

Q. 26 (a). please explain what happens in the above part of circuit

Reply
shivani sharma
12/30/2013 08:53:01 pm

sir ,in a parallel plate capacitor,when we give a charge +Q to one plate and it then induces a charge -Q on the inner surface of the capacitor and +Q on the outer surface and if the second plate is earthed,then its potential at the outer surface is 0,why we take the potential of the whole plate 0 as there is -Q on the inner surface ?
sir pl tell me as I am not able to understand this for a long time

Reply
shubhu
2/5/2014 03:32:16 am

Dear of second plate is earthed then whole of the induced +ve charges flows to earth to match the potential of earth which is taken to be zero. [ it is potential diff . Not just potential. Adding small amount of chrhe to earth doesn't alter its potential so pot. Diff. Is zero]
Now negative charge is still present on. Inner side which will tend to decrease the potential of main plate so that more of charge of charge can be stored .
Its principle of capacitor and I dun think any where its written that potential of whole of other plate ibecomes zero

Reply
theyatin
2/5/2014 09:26:36 pm

dear
good work.

shivani sharma
1/2/2014 08:21:42 pm

sir,pl can I get the reply

Reply
shivani sharma
1/4/2014 02:19:30 pm

sir,pl can I get the reply

Reply
shivani sharma
1/4/2014 09:22:01 pm

sir,pl can I get the reply

Reply
shubhu
2/6/2014 02:01:55 am

In Q 20 when the switch is closed so does the capacitor down it has nothing to do and we can then neglect it presence ?? Why is it so ? I thought as initially it was being charged so when switch is closed it has some stored charge due to which it is at higher potential then the wire holding switch!!
Need explanation!!
In q66 how the force equation is derived??

Reply
shubhu
2/6/2014 02:04:36 am

In Q 20 when the switch is closed so does the capacitor down it has nothing to do and we can then neglect it presence ?? Why is it so ? I thought as initially it was being charged so when switch is closed it has some stored charge due to which it is at higher potential then the wire holding switch!!
Need explanation!!
In q66 how the force equation is derived??

Reply
theyatin
2/10/2014 01:30:46 am

dear if close means connected.
then we can ignor as the capacitor may discharge in loop so we can ignor it.

Reply
Aradhana
2/9/2014 06:49:44 pm

Sir Q2 part b

Reply
Aradhana
2/9/2014 06:50:39 pm

Sorry sir, Q26 (b)

Reply
shivam
2/26/2014 07:20:35 pm

Sir Question 22 Pls ,Explain it in as detailed way as possible
h

Reply
shivam
2/26/2014 07:21:14 pm

Sir Question 22 Pls ,Explain it in as detailed way as possible
h

Reply
Admin
3/6/2014 04:04:24 pm

Please wait. Yatin sir will resume work within 2-3 days.

Reply
theyatin
3/7/2014 01:28:04 am

dear,
first we need to use formula of verticle distance covered by particle that is d1/2
now we can find E=V/R
also for capcitor A
V1
now total C from d1+d2
hence finding q
and putting its value in E
substituting value of E in verticla distance covered by projectile in first equation
we can find u^2

Reply
Help
4/23/2014 02:42:07 am

Sir can you help me on how to find equivalent capacitors for a circuit diagrram
The method they are doing i cant understand

Reply
sj
5/2/2014 01:32:31 am

I cont understand the sol. of que.(20),(44) Pls explain.

Reply
Er.pawan mirjaik
5/3/2014 05:40:07 am

dear its calculation mistake over there once you do caculation with
C= 5*10^-6. as 1 microfarrad = 1*10^-6 i mean to say that q = 50v
put these values in the given formula.you get same answer.

Reply
cp
5/2/2014 10:26:47 pm

Please explain ques. 55 of chapter 31 (capacitors).

Reply
ER. MJAIKIR
5/3/2014 06:33:46 am

Dear it should be noted that when a dielectric slab is introduced between the plates of the capacitor which is kept connected to the battery its
capacitance increases
potential difference between the plates remain unchanged
charge on the plates increases
and energy stored 1/2 CV^2 in the capacitor increase.
using this concept we use formula AS GIVEN IN THE SOLUTION MANUAL.
Keeping in mind the proper conversion of the values.
infact conversion is not done correct there so you just take care of it



Reply
@22
5/5/2014 12:38:05 am

Please explain about the force and its derivation used in ques. 66.

Reply
ER.MIRJAIK
5/5/2014 11:46:37 pm

ear,
as dielectric is attracted by electric field.
which is balanced by weight of block.
so putting force due to electric field equal to Mg we can have mass.
that's the formula of force exerted on the dielctric slab inserted between capacitor.

Reply
@22
5/5/2014 12:45:26 am

why in formula for common potential as used in question 47 (b) it is '+' for parallel and '-' for series connection .

Reply
shubhu
5/13/2014 05:16:29 pm

In paralle positive plate of one capacitor is connected to positive of other so sense of charge flow isl same for both its + and vice versa for series as terminal connected are opposite!!!

Reply
ambi
5/12/2014 09:00:26 pm

Cant understand question no.30 of capacitors

Reply
shubhu
5/12/2014 09:58:39 pm

Dear Q=CV so we have C , have to find V for that Q is needed . Q is the charge on inner plates that are equal and opposite!!
Now back to ques.. It says charge Q(2.10^-8) is given to positive plate and -Q/2 to the other plate so due to induction charges wil appear on innere plates !! So let charge q induces in one inner plate and -q on the other inner plate ! So different charges are [ Q-q] [q] [-q] [-Q/2+q] after that one have to find q and just q/C ! Find q using the fact that electric field inside metal plate is zero !! Refer pg138q-8

Reply
ambi
5/13/2014 01:06:27 am

Thank u very much this was really a help .but pls tell me for hcv will i have to folkow any other book like ncert ,rensick halliday etc.

ambi link
5/12/2014 09:03:35 pm

Reply
ER.MIRJAIKJ
5/13/2014 11:17:57 pm

D.C. PANDEY one of the useful book

Reply
ambi
5/12/2014 09:08:56 pm

What extra books we have to follow to solve a h c verma question ?also ans question no.1 short questions of capacitors

Reply
shubhu
5/13/2014 05:11:00 pm

This ques too has same process as U asked above !!
remembering that charge of capacitor is charge present on innere plates equal and opposite. So let q charge induces on inner plate of one and -q on the other so charge distribution is Q1-q , q, -q , -Q2+q again using the thought that electric field inside conductor is zero solve for q [ Q1+Q2]/2

Reply
Manish
5/14/2014 02:16:01 am

answer to ques 40 is wrong

Reply
AM
5/14/2014 09:08:59 pm

Can you pl explain that in Q 29 of Ch31 of hcv why do we consider the equivalent capacitance of the infinite ladder as the same as the unknown capacitance C?

Reply
ER.MIRJAIK
5/20/2014 02:13:10 am

It possess series arrangement with given capacitance

Reply
Vasudha
5/22/2014 03:54:38 am

question no. 33.....how are the 3 plates connected in series?? if they are then the potential differance between the first n last plate should be that of the battery that is also a condition for series ryt???

Reply
vandana
6/5/2014 05:33:39 am

can u please explain q 26.......what is star delta conversion

Reply
abhay
6/20/2014 02:45:07 am

please explain question 43. how was the formula for energy stored in front of plane derived?

Reply
Admin
6/22/2014 01:53:05 pm

Dear Students,
Our Teacher is unavailable for few days. We are sorry for this inconvenience. Our teacher will give your answers soon.

Reply
mruganshu
6/22/2014 02:24:47 pm

i have prob in q- 23 of exercise

Reply
shubham
6/23/2014 03:40:00 pm

I did not understood the solutipn of question number 63.

Reply
swadhin
6/24/2014 02:00:57 am

Sir plz explain me quest no 20 from exercise and 1,4 of short quest.

Reply
sumant
6/30/2014 03:31:24 pm

can u please explain q 26.......what is star delta conversion

Reply
ROSHAN link
7/1/2014 12:17:36 am

how did we get the charge in q38

Reply
Admin
7/2/2014 04:38:10 pm

Dear Students,
Our Teacher is unavailable for few days. We are sorry for this inconvenience. Our teacher will give your answers soon.

Reply
Febe
7/9/2014 12:51:45 am

Sir,can you please explain Q.26(b).explain star-delta conversion used in the problems also.

Reply
ER MIRAJANE (fairy tail)
7/24/2014 09:52:39 pm

sir can u pls tteellll meeee wats d rite answer of Q42 .......repliez vite

Reply
Natsu Dragneeell
7/24/2014 09:59:53 pm

sir kindly reply ASAP .....try 2 understand d situation ....GIMME D ANSWER OF q 42...D RITE 1 NOT D WRONG 1!!!!!!

Reply
ambi
8/30/2014 10:26:43 pm

well, take the smaller sphere of radius R and the inner part of the larger sphere 2R as a spherical capacitor since in the inner part of larger sphere charge induced by the smaller sphere will be same the charge of the inner sphere but of opposite sign and let the charge of smaller sphere be q and larger one on inner side be -q
so capacitance of this part becomes 8R.(pie).(epsilon 0)
now potential of the capacitor becomes
V'=1/[4*(pie).(epsilon 0)]{q/R-q/2r}
=1/[4*(pie).(epsilon 0)]{q/2R}
=1/[8*(pie).(epsilon 0){q/R}
now potential of only smaller sphere before 2R sphere was brought is
V=1/[4*(pie).(epsilon 0)]{q/R}
=2V'
=>V'=V/2
now enegy inside =energy of capacitor=
1/2*C*V'^2
=1/2 *8 (pie)(epsilon 0)*(V^2)/4
=(pie).(epsilon 0).V^2 (where * is multiplication and ^ is to the power)
b)now if we move a charge q from infinity to the surface of 2R sphere the amount of work done will be same as that moving the charge inside the sphere since elctric field inside the sphere is zero as charges inside a conducting surface is zero
in other words ,the potential energy of sphere remains same inside and outside of the sphere .

Reply
ambi
8/30/2014 10:26:53 pm

well, take the smaller sphere of radius R and the inner part of the larger sphere 2R as a spherical capacitor since in the inner part of larger sphere charge induced by the smaller sphere will be same the charge of the inner sphere but of opposite sign and let the charge of smaller sphere be q and larger one on inner side be -q
so capacitance of this part becomes 8R.(pie).(epsilon 0)
now potential of the capacitor becomes
V'=1/[4*(pie).(epsilon 0)]{q/R-q/2r}
=1/[4*(pie).(epsilon 0)]{q/2R}
=1/[8*(pie).(epsilon 0){q/R}
now potential of only smaller sphere before 2R sphere was brought is
V=1/[4*(pie).(epsilon 0)]{q/R}
=2V'
=>V'=V/2
now enegy inside =energy of capacitor=
1/2*C*V'^2
=1/2 *8 (pie)(epsilon 0)*(V^2)/4
=(pie).(epsilon 0).V^2 (where * is multiplication and ^ is to the power)
b)now if we move a charge q from infinity to the surface of 2R sphere the amount of work done will be same as that moving the charge inside the sphere since elctric field inside the sphere is zero as charges inside a conducting surface is zero
in other words ,the potential energy of sphere remains same inside and outside of the sphere .

Reply
shushant
12/4/2014 07:35:51 pm

i have problm wth the question no. 38 how did u find q there???

Reply
Apurv Tiwari
1/11/2015 09:42:23 pm

I want to ask that in last question, if we take the displacement from one extreme to mean position, then it would be

s= l-a. Then the time period comes out to be the one that is given in the book. But if e use this method which is used here, then we get a slightly different time period with root8 instead of 8 as in the above method. So, which method should we use?

Reply
ankit singh
2/23/2015 03:00:28 am

want some ideas about common potential

Reply
redad
4/1/2015 04:59:54 pm

sir please help me with the solution of this question
The gap between the plates of a parallel
plate capacitor of area A and distance
between plates d, is filled with a dielectric
whose permittivity varies linearly from e1
at one plate to e2 at the other. The
capacitance of capacitor is :

Reply
saral
4/7/2015 01:58:32 am

I am having problem with q20

Reply
shubhhi
5/3/2015 10:19:47 pm

pls help me with que 57 of exercise.

Reply
kunal
5/14/2015 03:05:14 am

I have prblm in ques26(a) plz help me

Reply
Rahul
5/14/2015 01:43:40 pm

I didn't get the question no. 8
The equivalent capacitance came 4
But why did we multiply the capacitances by 6?

Reply
Tanya
5/23/2015 09:45:24 pm

Please explain question no. 34 in brief.

Reply
Rohit
6/9/2015 03:10:10 am

Please derive the force equation usedin Q 66

Reply
Ayush
7/14/2015 12:58:41 pm

Sir i have problem in the solution of question no.57.
I didn't get how it is integrated.

Reply
Amol link
7/29/2015 02:03:36 pm

sir i don't understand when dielectrics are in series and when they are parallel

Reply
Amol link
7/29/2015 02:10:20 pm

sir i didn't understood why in q 63,62 of capacitors you took spherical capacitors in series

Reply
akanksha mishra link
8/6/2015 03:29:56 am

this helps a lot in studying. Plz keep some extra problems out of book.

Reply
Rachana Rudra
8/14/2015 01:34:01 am

i didnot understand why q=4x10^-4 in question no 38.........it is not even given in the question but in the solution it is there....& i cannot calculate it out also??????

Reply
Abhishek Mishra
8/19/2015 09:20:11 am

Reply
Anamika Bhardwaj
5/8/2016 07:23:27 am

On question 20 why does c eff after switch is closed become 10?

Reply
KRITGYAA
6/6/2016 12:45:53 am

I DIDNT GET 25 (D) CAPACITOR

Reply
Mugdha
6/12/2016 09:11:56 am

Plz explain sol 19

Reply
Yatharth
6/12/2016 09:13:32 am

Please explain sol 20

Reply
Mrinz link
6/18/2016 10:06:07 pm

Tell me question 4 objectives of hc verma capacitors

Reply
xitiz
6/25/2016 10:19:32 pm

sir i want to know ques no 3 diagram of objective 1 of capacitor

Reply
LEO TOLSTOY
7/2/2016 07:44:54 am

HCVERMA IS THE BEST BOOK EVER SEEN BY ME IN MY LIFE OF M...A...N...Y... YEARS

Reply
piyush
7/3/2016 08:48:43 pm

i have a problem on question no. 20 capacitor

Reply
Ashish
7/28/2016 08:58:34 am

Chapter 31 question no 25 b.there is a battery. Of 12 volt each...but in solution , it is taken as two battery of 12 and 24 volt...is there any printing mistake in the book

Reply
Manish
10/2/2016 01:09:40 am

Sir i want to improve my 3d thinking ability for solving physics problem. I leaen and understand every concept easily but when i am going to application i am little bit confused so pls help me for developing my ability pls sir reply as soon as possible. I am waiting for your answer

Reply
monisha
5/25/2017 09:25:44 pm

sir I am having problem in objectives of capacitor question 1 and 8 of objective 1 and question 4,5,6 of objective 2

Reply
monisha
5/27/2017 04:47:15 am

sir I am having problem in objectives of capacitor question 1 and 8 of objective 1 and question 4,5,6 of objective 2

Reply
Rohit
4/6/2018 12:23:08 pm

Answer of q.no.59 is wrong in your solution

Reply
Anonymous
6/22/2018 01:37:31 am

Please explain the answer to question 50 c part

Reply



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