Common potential and loss in energy on shearing charges .
Before to know about the topic Common potential and loss in energy on shearing charges .student must have to know about Energy stored in the capacitor and electrostatic potential .
In this topic students will know about Common potential and loss in energy on shearing charges . its derivation and numerical is important for class 12th ( Board examination) and for NEET/JEE examinations .
Common potential – When two charged capacitors are connected with each other using a conducting wire then charge starts flowing from high potential to low potential, and after some time their potential becomes equal.
Let two capacitors C1 and C2 and their potentials are V1 And V2 respectively. Then total charge before combination Q = C1V1 + C2V2.
Let they are connected with a conducting wire, then charge start flowing from high potential to low potential, till potential is not equal. Let V is the common potential then total charge after combination Q = V (C1 + C2).
But from the conservation of energy charge after collision = charge before collision
So, C1V1 + C2V2 = (C1 + C2) V
Then V = C1V1 + C2V2 / (C1 + C2)………………………eq.
Loss of energy on shearing charges –
As we have seen in the above equation common potential after combination
V = C1V1 + C2V2 / (C1 + C2)
Total energy before combination U1 = ½ C1V12 + ½ C2V22
Total energy after combination U2= ½ ( C1 + C2 ) V2 ; Putting the value of V
We get , U2= ½ ( C1 + C2 ) [C1V1 + C2V2 / (C1 + C2)]2
= ( C1V1+C2V2 )2 /2(C1+C2 )
Then loss in energy ;
=U1-U2 = [½ C1V12 + ½ C2V22] – [( C1V1+C2V2 )2 /2(C1+C2 )]
= [C1V12(C1+C2 )+C2V22 (C1+C2 )- (C1V1+C2V2)2]/2(C1+C2) ;
= C1C2(V12+V22-2V1V2)/2(C1+C2) Or U1-U2 = C1C2(V1-V2)2/2(C1+C2) . Which is positive for all possible values of C1 , C2 And V1 , V2 .
I.e . U1 – U2 >0 ( POSITIVE ) .
So, U1 > U2 . Therefore we can say that there is always loss in energy during the combination of charged capacitors .