Electrostatic potential using method of imagesby tomwilliam2 Tags: electrostatic, images, method, potential 

#1
Sep2212, 03:03 PM

P: 64

1. The problem statement, all variables and given/known data
There is a charge q, at a distance d from an infinite conducting plane (z=0). Determine the electrostatic potential drop between the z=0 and z=d. 2. Relevant equations Ohm's Law DeltaV=integral (E.dl) 3. The attempt at a solution I know how to do this problem, using J = sigma E, Ohm's law, model the situation with a charge q at z=d, find the electric field, then integrate over the line to find an expression for the potential function V(z). I know I can integrate this potential function over the line now to find the potential drop, using the boundary condition that V=0 on the conducting plane. The one thing I don't understand is the units. If I have a potential in Volts, and I integrate over a line, how can it result in a potential drop? Wouldn't it be V m? Thanks in advance 



#2
Sep2212, 05:51 PM

HW Helper
P: 5,004

Wouldn't the potential drop (or difference) just be V(d)V(0)? Integrating the electrostatic potential V(z) from z=0 to z=d, would give you the change in electrostatic energy (sometimes confusingly referred to as electrostatic potential energy).




#3
Sep2312, 03:24 AM

P: 64

I thought that too, but from the worked solutions I get the potential as:
$$V(z) = \frac{I}{4\pi \sigma z}  \frac{I}{4\pi \sigma (2d  z)}$$ Where sigma is conductivity. It goes on to say: "by integrating this expression between the real source and the metal boundary, we could obtain an expression for the potential drop deltaV, and dividing this by the currnt would give resistance." I thought it was a mistake, but this online source says the same (para three): http://physicspages.com/2011/10/08/c...lectrostatics/ Is this all just wrong? 



#4
Sep2312, 03:48 AM

HW Helper
P: 5,004

Electrostatic potential using method of imagesIn any case, the beginning of the statement "by integrating this expression" is ambiguous without context that allows one to determine which expression they are talking about integrating. I suspect it is the electric field they are integrating to get the potential drop or difference. [tex]V(b)  V(a) = \int_a^b \mathbf{E} \cdot d\mathbf{l}[/tex] 



#5
Sep2312, 02:32 PM

P: 64

Thanks for the help.
I think my textbook has a mistake in it...that's why when it uses the final expression to calculate resistance it comes out in Ohm metres. It was worth going through the worked example with a fine toothcomb after all! 


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