super awesome.. this video proved that there is a difference between having knowledge and teaching ... now I see how lame most of the teachers are..sir you are GOD of Physics
Sir if the dielectric is inserted at velocity v , then at time t to find the capacitance, should I find the amount inserted and then do the case of two parallel capacitors?
Because in second case battery is disconnected so no return path is available for charge to get discharged thats why in an isolated/disconnected capacitor charge remain constant...
Electric field will change sir....wouldn't it?....as d will change....as force between the capacitor plates will change ... and hence the force must change.....please sir clear my doubt
Capacitance does not depend upon field or potential difference its q/V and here charge will increase to K times... or you directly use the formula of capacitance = ε0A/d so here ε increases K times and A, d remain same...
No as C = ε0A/d when no dielectric and it becomes Kε0A/d so C becomes KC now so final charge also becomes KCV from CV as V is constantly maintained here...
yes but in the expression for C it shows that it does not depend on Q or V as V is directly proportional to Q so C is a constant which depends only upon the dimensions and medium characteristics...
Could you please make a video on the following. A dielectric is released into a capacitor [ with ; without battery ]. What will be its KE , heat produced in connecting wires , heat produced in slab, energy spent in polarizing the dielectric. Also is there any difference if the dielectric is just released vs slowly released. Is heat produced same in both ? Thanks.
Already there in advance illustrations... In case of dielectric polarization, heat is not analyzed or generally neglected as that goes beyond the scope of these lectures... at high school physics you need to ignore heat released in case of slowly moving dielectrics in capacitor plates... In case of high kinetic energy heat cannot be ignored and not considered here...
watch the next video in playlist or click on the link below... www.physicsgalaxy.com/lectures/1/5/24/1071/Force-on-a-Dielectric-Slab-During-Insertion-in-a-Capacitor
super awesome.. this video proved that there is a difference between having knowledge and teaching ... now I see how lame most of the teachers are..sir you are GOD of Physics
Very good illustration... Helped a lot..... Sir pls make few last moment tips for NEET and other exams
Yes will do...
Sir if the dielectric is inserted at velocity v , then at time t to find the capacitance, should I find the amount inserted and then do the case of two parallel capacitors?
Excellent video thank you!
Best explanation I have ever got🙏
Thanku so much sir🙏
Sir one video is not in the playlist of Capacitance.
Earlier there were 40 videos and now only 39.
Please upload that video.
Sir how is the charge not changing in second case?
As q=CV in which C is changing so why will it not make any difference plz elaborate.
Because in second case battery is disconnected so no return path is available for charge to get discharged thats why in an isolated/disconnected capacitor charge remain constant...
Electric field will change sir....wouldn't it?....as d will change....as force between the capacitor plates will change ... and hence the force must change.....please sir clear my doubt
Wow, great explanation of a trick concept
concept now clear sir
Thank sir I could complete my assignment cause of u r video thank you very much sir
Sir the electric field remains the same
So how can the energy change?please reply since energy is energy density into volume
The work done by external agent to insert the dielectric results in decrease in capacitor energy change...
W(in inserting the dielectric)=|U-U/k|...
Sir when to use Electric field= Q/AEo??
Thank you sir!!!
Where does the energy go in the second case.... As energy becomes u by k times
Sir,thinking practically ,how can insertion of dielectric when battery isn't present alter the energy available to it i.e. reduce it by k times?
I have same doubt
Because the electric energy is consumed in Work done to insert the dielectric inside the capacitor
Although I replied 4 years later😂
Sir,Rc circuit capacitance me cover nahi
What will happen to the force between plates sir
sir,please can i get the notes for this
has sir given anything ???i also asked in the other videos but no answer yet.....please tell me as soon as sir tells anything thanks man appreciated
@@adarshkundu5995 You can buy Physics galaxy lecture notes on Amazon
Sir,in the first case why capacitance increases K times even though field and potential remain same??
Capacitance does not depend upon field or potential difference its q/V and here charge will increase to K times... or you directly use the formula of capacitance = ε0A/d so here ε increases K times and A, d remain same...
Sir,why charge increases K times?Initially there was CV charge,on capacitor plates after inserting dielectric also charge can remain same on plates??
No as C = ε0A/d when no dielectric and it becomes Kε0A/d so C becomes KC now so final charge also becomes KCV from CV as V is constantly maintained here...
Sir but we got C = ε0A/d from C=Q/V,so while deriving from definition viz.C=Q/V we know potential is not changing,how to decide about charge?
yes but in the expression for C it shows that it does not depend on Q or V as V is directly proportional to Q so C is a constant which depends only upon the dimensions and medium characteristics...
Could you please make a video on the following. A dielectric is released into a capacitor
[ with ; without battery ]. What will be its KE , heat produced in connecting wires , heat produced in slab, energy spent in polarizing the dielectric. Also is there any difference if the dielectric is just released vs slowly released. Is heat produced same in both ?
Thanks.
Already there in advance illustrations... In case of dielectric polarization, heat is not analyzed or generally neglected as that goes beyond the scope of these lectures... at high school physics you need to ignore heat released in case of slowly moving dielectrics in capacitor plates... In case of high kinetic energy heat cannot be ignored and not considered here...
Sir how to get force on dielectric.
watch the next video in playlist or click on the link below...
www.physicsgalaxy.com/lectures/1/5/24/1071/Force-on-a-Dielectric-Slab-During-Insertion-in-a-Capacitor
sir will the charge on the capacitor decrease due to induced charges on the dielectric in the 2nd case
In case II as battery is not there so charge on capacitor will remain same...
sir i think
U changes like U/k
In Case-I it is kU and in case II if U/k watch the video again and patiently try to understand it...