The soulution in the HC Verma is different from what I have found using capacitors in series and parallel. why is there incosistency?

Asked by arss9101127 | 11th May, 2020, 08:56: AM

Expert Answer:

Find the equivalent capacitance between point A and B (31-W5a).  
 
Solution - 
 
Let's consider that the battery is connected between the points A and B. Cathode of battery supplied charge - Q while anode is source of + Q charges. 
 
T h e space c h a r g e space Q space i s space d i v i d e d space b e t w e e n space t h e space p l a t e s space o f space a n space c a p a c i t o r space A space c h a r g e space Q subscript 1 space i s space o n space t h e space p l a t e space a space a n d space r e s t space left parenthesis Q minus Q subscript 1 right parenthesis space o n space C subscript 2 space p l a t e space e space T h e n space b y space s y m m e t r y space colon space minus space C h a r g e space o n space p l a t e space g space i s space Q subscript 1 space a n d space left parenthesis Q minus Q subscript 1 right parenthesis space o n space t h e space c space T h u s comma space c h a r g e space o n space C subscript 3 space w i l l space b e space left parenthesis 2 Q subscript 1 space minus space Q right parenthesis space o n space p l a t e space i space V subscript A space minus space V subscript B space equals space left parenthesis V subscript A space minus V subscript D right parenthesis space plus space open parentheses V subscript D space minus V subscript B close parentheses V subscript A space minus space V subscript B space equals space fraction numerator Q subscript 1 over denominator C 1 end fraction plus fraction numerator Q minus Q subscript 1 over denominator C 2 end fraction space... open parentheses A s comma space V space equals Q divided by C close parentheses space... space left parenthesis 1 right parenthesis space N o w thin space w e thin space c a n thin space w r i t e comma V subscript A space minus space V subscript B space equals space left parenthesis V subscript A space minus V subscript D right parenthesis space plus open parentheses V subscript D space minus V subscript E close parentheses space plus thin space left parenthesis V subscript E space minus space V subscript B right parenthesis V subscript A space minus space V subscript B space equals space Q subscript 1 over C subscript 1 space plus space fraction numerator 2 Q subscript 1 space minus Q over denominator C 3 end fraction space plus fraction numerator Q subscript 1 over denominator C 1 end fraction space... left parenthesis 2 right parenthesis T h u s comma space e q n space left parenthesis 1 right parenthesis space c a n thin space b e thin space w r i t t e n thin space a s comma space V subscript A space minus space V subscript B space equals space Q subscript 1 open parentheses fraction numerator 1 over denominator C 1 end fraction minus fraction numerator 1 over denominator C 2 end fraction close parentheses plus fraction numerator Q over denominator C 2 end fraction open parentheses fraction numerator C subscript 1 C subscript 2 over denominator C subscript 2 minus C subscript 1 end fraction close parentheses open parentheses V subscript A space minus space V subscript B close parentheses space equals space Q subscript 1 plus open parentheses fraction numerator C 1 over denominator C subscript 2 minus C subscript 1 end fraction close parentheses Q space space space... left parenthesis 3 right parenthesis E q thin space left parenthesis 2 right parenthesis space b e c o m e s comma space V subscript A space minus space V subscript B space equals space 2 Q subscript 1 open parentheses 1 over C subscript 1 plus 1 over C subscript 3 close parentheses minus Q over C subscript 3 space open parentheses fraction numerator C subscript 1 C subscript 3 over denominator 2 open parentheses C subscript 1 plus C 3 close parentheses end fraction close parentheses open parentheses V subscript A space minus space V subscript B close parentheses space equals space Q subscript 1 minus open parentheses fraction numerator C subscript 1 over denominator 2 open parentheses C subscript 1 plus C subscript 3 close parentheses end fraction close parentheses Q space space... left parenthesis 4 right parenthesis s u b t r a c t i n g space left parenthesis 4 right parenthesis space f r o m space left parenthesis 3 right parenthesis space w e space g e t comma space open parentheses V subscript A space minus space V subscript B close parentheses space open square brackets open parentheses fraction numerator C subscript 1 C subscript 2 over denominator C subscript 2 minus C subscript 1 end fraction close parentheses minus open parentheses fraction numerator C subscript 1 C subscript 3 over denominator 2 open parentheses C subscript 1 plus C subscript 3 close parentheses end fraction close parentheses close square brackets equals open square brackets open parentheses fraction numerator C subscript 1 over denominator C subscript 2 minus C subscript 1 end fraction close parentheses plus open parentheses fraction numerator C subscript 1 over denominator 2 open parentheses C subscript 1 plus C subscript 3 close parentheses end fraction close parentheses close square brackets Q space open parentheses V subscript A space minus space V subscript B close parentheses space left parenthesis 2 C subscript 1 C subscript 2 space left parenthesis C subscript 1 plus C subscript 3 right parenthesis space minus space C subscript 1 C subscript 3 open parentheses C subscript 2 minus C subscript 1 close parentheses space right parenthesis equals C subscript 1 left parenthesis 2 left parenthesis C subscript 1 plus C subscript 3 right parenthesis plus left parenthesis C subscript 2 minus C subscript 1 right parenthesis right parenthesis Q C subscript e q end subscript space equals space Q over V equals fraction numerator Q over denominator open parentheses V subscript A space minus space V subscript B close parentheses end fraction equals open square brackets fraction numerator 2 C subscript 1 C subscript 2 plus C subscript 2 C subscript 3 plus C subscript 3 C subscript 1 over denominator C subscript 1 plus C subscript 2 plus 2 C subscript 3 end fraction close square brackets  
Please compare your answer with this solution as this is the ideal method to solve such numericals. 

Answered by Shiwani Sawant | 11th May, 2020, 10:16: PM

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