Calculating Work to Charge Parallel Conducting Plates

AI Thread Summary
To calculate the work required to charge a pair of parallel conducting plates with charges of +1.0 x 10^-6 C and -1.0 x 10^-6 C, the relevant equations are W = qV and V = Qd/(ε₀A). The user initially obtained a result that was double that of the textbook, prompting confusion regarding the electric field of a single plate and the initial state of the plates. It was pointed out that the voltage changes as the plates are charged, necessitating the use of integration for accurate calculation. Understanding the dynamics of charging is crucial for solving this problem correctly.
razidan
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Homework Statement


A pair of parallel conducting plates, each measuring 30 cm X 30 cm, are separated by a gap of 1.0 mm. How much work must you do against the electric forces to charge these plates with ##+1.0 \cdot 10^{-6} C ## and ##-1.0 \cdot 10^{-6} C##, respectively?

Physics for Engineers and Scientists (Third Edition), Ohanian and Markert, Vol 2. pp 823. Q.55

Homework Equations


##W=qV##
##V_{plates}=\frac{Qd}{\varepsilon_0 A}##

The Attempt at a Solution


So, plugging in the numbers gives a result twice the result in the book.
This leads me to think about a field of a single plate, ##E=\frac{Q}{ 2 \varepsilon_0 A}##, but I can't figure out why. and more in general, I'm not sure what the question is describing - are the plates initially neutral. and then they are charged?

Thanks,
R.
 
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You didn't take into account the fact that V is changing while you charge up the plate. You must integrate ##V \:dq##
 
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Gene Naden said:
You didn't take into account the fact that V is changing while you charge up the plate. You must integrate ##V \:dq##
Oh boy...
Thanks.
 
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