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An air -cored capacitor of plate area A and separation d has a capacity C . Two dielectric slabs are inserted between its plates in two different manners as shown . Calculate the capacitance in each case . |
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Answer» Solution :(i) Let the charges on the are Q and -Q. Electric FIELD in free space is `E_(0)=(sigma)/(epsilon_(0))=(Q)/(Aepsilon_(0))` Electric field in first SLAB is `E_(1) =(E_(0))/(K_(1))=(Q)/(Aepsilon_(0)K_(1))` Electric field in second slab is `E_(2)=(E_(0))/(K_(2))= (Q)/(Aepsilon_(0)K_(2))` The POTENTIAL difference between the plates is `V = E_(0)(d-t_(1)-t_(2))+E_(1)t_(1)+E_(2)t_(2)` `implies V=E_(0)(d-t_(1)-t_(2)+(t_(1))/(K_(1))+(t_(2))/(K_(2)))"" ("As"E_(1)=(E_(0))/(K_(1)).E_(2)=(E_(0))/(K_(2)))` `:. V=(Q)/(Aepsilon_(0))(d-t_(1)-t_(2)+(t_(1))/(K_(1))+(t_(2))/(K_(2)))` `:.C=(epsilon_(0)A)/(d-t_(1)=t_(2)+(t_(1))/(K_(1))+(t_(2))/(K_(2)))` (ii) The capacitor can be CONSIDERED as TWO capacitors of capacitances `C_(1)` and `C_(2)` in parallel .  `implies C_(1)=K_(1) (epsilon_(0)A_(1))/(d). C_(2)=(K_(2)epsilon_(0)A_(2))/(d)` `implies ` Total capacitance `C= C_(1)+C_(2)` `implies C=((K_(1)A_(1)+K_(2)A_(2))epsilon_(0))/(d)` |
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