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

AI Thread Summary
The discussion focuses on calculating the electric field, charge storage, and energy storage of a 9V battery connected to two parallel plates separated by 3mm. The electric field magnitude can be determined using the formula E = V/d, resulting in 3000 V/m. The charge stored can be calculated with the capacitance formula C = Q/V, leading to a charge of 0.015 C for the given plate area of 0.5 m². Energy storage is found using the formula for energy in a capacitor, yielding 0.0675 Joules. These calculations illustrate fundamental principles of capacitors and electric fields.
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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 postive 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 \sigma = Q / A.

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

Energy stored in a capacitor is \int_0^V q dv.

This should get you started.
 
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