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Charge upon a parallel plate capacitor
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[QUOTE="Pushoam, post: 6852809, member: 619344"] [ATTACH type="full" align="left" width="209px"]321970[/ATTACH] Let's say that the battery capacitor has capacitance C[SUB]b[/SUB], initial charge Q[SUB]b[/SUB], and potential difference V[SUB]b[/SUB] and the capacitor has charge Q on each plate. After the connection and coming to equilibrium, the system has the charge configuration given in the figure and potential difference V'[SUB]b[/SUB]. Calculations similar to those in post #9 show that outer surface charges on the plates of battery are same. Now, $$ q_b = C_b V'_b ~~\text{and}~~q = CV'_b$$ $$\Rightarrow q = \frac {q_b C}{C_b} ~~\text{and} ~~q_b = \frac{Q_B}{1+\frac C {C_B}}$$ So, after the connection, potential difference and facing surface charges of each plate changes. The charge Q, too, redistributes on the outer surface of all plates as the connecting plates of the battery and capacitor are equipotential. Could you help me in finding the distribution? On the other hand, an ideal battery is supposed to maintain the potential difference. Here, the potential difference changes. So, the capacitor battery is not an ideal battery. [/QUOTE]
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Introductory Physics Homework Help
Charge upon a parallel plate capacitor
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