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(a) Describe briefly the process of transferring charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor. (b) A parallel plate capacitor is by a battery to a potential difference V. It is disconneted from battery and then connected to another uncharged capacitor of the same capaitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor. |
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Answer» Solution :(a) If a parallel plate capaccitor is connected across a battery, the electric charge will flow thrugh the circuit. As the charges reach the plate, the insulating gap does not allow the charges to MOVE further, hence positive charge get diposited on the SIDE of the plate which is connected to the positive terminal of the battery and negative charge on the plate which is conneted to the negative terminal of the battery. Thus the current thrugh the circuit gradually becomed less and become zero and POTENTIAL difference between the plates of the CAPACITOR become equaland opposite to the battery. Energy Stored in a Parallel plate Capacitor : When a capacitor is charge, work is done. This work is stored as electrical potantial energy. Suppose a capacitory is charged with a charge q. Potential difference across the plates of the capacitor `V=-q/C` Work done to increse the small amount of charge `dW=Vdq=q/Cdq` Total work doen, `W=1/Coverset(O)underset(0)(int)qdq` `W=1/Coverset(Q)underset(0)([q^(2)/2])=1/2Q^(2)/C` `"Potential Energy,"U=1/2Q^(2)/C=1/2CV^(2)=1/2QV` Let, the charge on the first capacifor be q, Potantial difference be V and Capacitance be C `"Initial Potantial Energy "=U_(1)=1/2CV^(2)` `"Common Potential of the CombinationV'"=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))` `"as "V_(2)=0" and "C_(1)=C_(2)=C""V'=(CV)/(2C)=V/2` `"Net Capacitance of the Combination, "C'=C_(1)+C_(2)=2C` `"Final Potantial Energy."U_(f)=1/2C'V'^(2)` `U_(f)=1/2(2C)(V/2)^(2)=(CV^(2))/4` `"Dividing equation (1) by equation (2) "U_(f)/U_(i)=(CV^(2))/42""2/(CV^(2))=1/2` Hence, `U_(f):U_(i)=1:2` |
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