11ya c my soln..i mentioned it will be 2r but took it as r myself in calculation..small mistake..otherwise i think its ok..
u check it too once n tell if that calculation mistake is d reason final ans is not matching.
Q1) when a dielectric slab is inserted between the plates of a parallel plate capacitor the energy of the system decreases.What can be said about the force on the slab exertd by the feild?
Q2) A positive charge of 2*10^-8 coulumb is placed on the positive plate of a capacitor and a -ve charge of -1*10^-8 Coulumb ont he -ve plate of a capacitor.Calculate the potential diff developed between the plates of the capacitor?
Q3) A metal sphere of radius r is charged to a potrntial v .Find the electrostatic energy stored in the feild within a radius of 2r.
Q4) a CAPACITOR HAVING CAPACITENCE 100 MICRO fARAD is charged to a potential difference of 24 V.. it is then reconnected to another battery of emf 12V after disconnecting the original batery. Find the charge ont he plate after reconnection.and also find the heat developd during the reconnection....
Q5)
Q6) 
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19 Answers
11Ronald... when u say three capacitors are in series won't the charge on each capacitor be the same ?
1111Ronald Very sorry for the late reply...
The answer to question 3 is ΠεR2V
11wont it simply be 2-1 *10^8??? = 1*10^-8
i checked the answer from HCV for Question 3
11@ GNB - yes 3 cud b wrong..though i am quite sure its ok..we'l wait till aditya confirms..
in 2 i subtracted to find net charge,,just because one is a pos charge n other neg...
11in q 2 why have u taken net charge as q1-q2/2? wont it simply be q1+q2???
11@ decoder..
capacitance b/w plates has to be given..he probably forgot..i just checked it up in HCV,,they took that c value in their sum.
114.)
After connection
When C = 100 μf V = 12 V
q = CV = 1200 μc (After connection)ï€ (Ans)
After reconnection
C = 100 μc, V = 12 v
The energy appeared = (1/2) CV2 = (1/2) × 100 × 144 = 7200 J
This amount of energy is developed as heat. (Ans)
113.)
Q = CV = 4Ï€EorV
E=q2/2c
u know C= 4Ï€Eor (here it will be 2r)
therefore E= 2Ï€Eorv2 (Ans)
112.)
q1 = +2.0 × 10–8 c
q2 = –1.0 × 10–8 c
C = 1.2 × 10–3 μF = 1.2 × 10–9 F
net q = (q1 - q2)/2
v=q/c = 12.5 V (ans)
115 is easy see..
draw the fig...give the plates their charges..u c there are 3 capacitances formed.. ( u got it ??)
A given..d given..
c=EoA/d..... that is coming (24 × 10–9 F).
as 3 capacitances are in series..
Ceq= C/3 = 8 × 10–9
=> The total charge to a capacitor = 8 × 10–9 × 10 = 8 × 10–8 c (q=cv)
The charge of a single Plate = 2 × 8 × 10–8 = 16 × 10–8
= 0.16 × 10–6
or 0.16c.