Electric Field and Energy Storage of 9V Battery Across Parallel Plates

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SUMMARY

The discussion focuses on calculating the electric field, charge storage, and energy storage of a 9V battery connected across two parallel plates separated by 3mm. The electric field (E) can be calculated using the formula E = V/d, resulting in an electric field of 3000 V/m. The capacitance (C) is determined using C = Q/V, where the area of the plates is 0.5 m², leading to a charge storage of 0.015 C. The energy (U) stored in the capacitor is calculated using U = 0.5 * C * V², yielding an energy storage of 0.0675 Joules.

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A 9v battery is connected across 2 parallel plates separated by 3mm.
a. what is the magnitude of the electric field between them?
b. the plates have an area of .5m^2, how much charge can be stored?
c. how much energy can be stored?

please help me with this question. i am lost and have no idea what formulas to use.
 
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Have done a chapter on capacitors? These questions use very standard formulae. If not, why don't you try to derive the formulae? Let me give you a headstart on the derivations

Capacitance C of a capacitor is defined as C = Q / V, where Q is the positive charged stored on the plates (not the total charge, just the charge on a single plate. Note that the total charge on a capacitor is always Q + (-Q) = 0) and V is the potential difference b/w the two plates.

Given a plate with area A and charge Q, its surface density of charge is [itex]\sigma[/itex] = Q / A.

For any infinite planar charge sheet with surface density [itex]\sigma[/itex], the electric field perpendicular to the sheet is [itex]\texbf{E} = \frac{\sigma}{2\epsilon_0}[/itex].

Energy stored in a capacitor is [itex]\int_0^V q dv[/itex].

This should get you started.
 

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