Q29

Q41,Q44(b)

Q26(b) (d)

For question 29

that section has always capacitance =C

That means

\displaystyle C=\frac{1}{\frac{1}{4}+\frac{1}{C}}+2

2 Likes

Q 41.Take that distance very close to surface then you get that E=\frac{1}{2}CV^2

C=4π\epsilon_0R,V=\frac{Q}{4π\epsilon_0R}

1 Like

For 26 use nodal analysis

For 45 intial charge =CV_1,V_1=24V

For final charge C(V_1+V_2) ,V_2=12V

Final charge should be \ CV_2

1 Like

Ya My mistake

1 Like

If the ques 41 has asked electrostatic energy inside the sphere then what would be the ans?

It's easier if you have studied potential energy of solid and hollow sphere in conductors....

BTW, if they would have asked for energy inside sphere it would be kq²/10r.

Can you tell me the source from where i can study these

I am sorry i think I messesed up and mixed two concepts, you can still study it though...

1 Like

Q48. Please , i have doubt here that since both capacitor and battery are at same potential so charges should not flow?

and in ques 52 i have some doubt please explain that as dielectric is placed in capacitor its potential become V/k , but the battery provide charge till whole capacitor gets charged so i mean that battery will provide charge till both capacitor and battery are at same potential ,so potential should become V of capacitor as capacitor is a charge storing device and it will store till its potential become equal to batteryQ53 (d) whole ques is how much charge has been flown through the battery after slab is inserted ?

As here battery is connected so only capacitance changes not potential

It you dispatch the battery then the charges will rearrange such that potential become \frac{V}{k}

1 Like

Bro but both capacitor and battery are at same potential so how can charges flow?

The positive plate was at 12V

But after connecting it was at (-12V)

ΔV=24V

2 Likes