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Introductory Physics Homework Help
Charge upon a parallel plate capacitor
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[QUOTE="Pushoam, post: 6852500, member: 619344"] [ATTACH type="full" align="left" width="201px" alt="1675849090192.png"]321947[/ATTACH] Taking the brown cylindrical surface as the Gaussian surface and applying the Gauss's law and considering the fact that electric field inside the material of conductor is 0 gives that surface charge densities on the facing surfaces are equal and opposite. Hence, the charges on the facing surfaces are q and -q. Now, for outer surface, Electric field at P is calculated first using the Gaussian surface and applying Gauss's law and then, due to all the charged surfaces individually and then the two expressions are equated. $$ \frac {q'}{A\epsilon_0} = \frac {q'}{2A\epsilon_0} +\frac {q}{2A\epsilon_0} +\frac {-q}{2A\epsilon_0} +\frac {q"}{2A\epsilon_0} $$ $$q' =q" $$ Hence, the inner surface charges of both plates are equal and opposite and the outer surface charges of both plates are equal and same. The battery doesn't supply the charges. It creates the potential difference across the plates. As a result, electrons of plate at higher potential move towards plate at lower potential creating inner surface charge q and -q. Now, the conservation of charge gives, $$ q' +q -q+q' = 2Q \Rightarrow q' =Q$$ Hence, the correct options are a, c and d. [/QUOTE]
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Charge upon a parallel plate capacitor
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