Electric Field, Potential Between 3 Charged Electrodes

AI Thread Summary
The discussion revolves around calculating the electric field (E) and potential (V) between three charged electrodes. The initial calculations suggested E as 1.4 x 10^7 V/m and V as -1.4 x 10^5 V, but the answer key indicated E as 1.4 x 10^8 V/m and V as 140,000 V, leading to confusion about the order of magnitude. An alternative method involving the calculation of separate surface charge densities (sigma) for each electrode was mentioned, raising questions about whether this approach is valid when the charges differ. The potential difference calculated using capacitance also aligned with the book's voltage, suggesting a possible error in the answer key's values. Overall, the discussion highlights the complexities in electric field calculations involving multiple charged electrodes.
Bogus_Roads
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Homework Statement



Capture.PNG


The book also asks for E versus x.


Homework Equations



dV/dx=-E
E=sigma/epsilon0

The Attempt at a Solution



This question seemed fairly straightforward: I assumed it could be treated like two +-50 nC capacitors, each with d=1 cm. According to my calculations, (from x=2 cm to x=3 cm for example) E would be 1.4 x 10^7 V/m, and V would be -Es=-1.4 x 10^5 V. The answer key says that E=+- 1.4 x 10^8 V/m, and that V= 140,000 V... I can't figure out how they got E to be 10^8, and if so, what it could have been multiplied by to get 1.4 x 10^5.

I googled it and found someone give an alternate solution in which he/she calculated separate sigma for the three electrodes (50 nC=Q for the first, 100 nC=Q for the second), and then added them and divided by epsilon0 to get the electric field...I didn't know you could do this; is it still a capacitor then, if the charges are different on either plate..? In any case, this answer was still nothing near the book's, off by a factor of 10^5 or something.

Thanks
 
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Looking at the answer key values, if E = 1.4 x 108 V/m and the separation between plates is 1cm, then that would make the potential difference between plates 1.4 x 106 V, which doesn't match their given voltage value. So perhaps they've mucked up the order of magnitude on the field strength.

If you do treat an adjacent pair of plates as a capacitor, then the plate area is (2cm)2 and separation 1cm, giving a capacitance of 0.354 pF. With a charge of 50nc, that yields a voltage of 1.41 x 105V, which does match their given voltage.
 
Never mind!
 
Thanks gneill,

So 1.4 x 10^5 V would make 1.4 x 10^7 V/m correct?

Does it make sense to add values of sigma to find the electric field, as was done in the Cramster problem I found online?
 
Maybe they used millimeters instead of centimeters for the distance between the plates.
 
Bogus_Roads said:
Thanks gneill,

So 1.4 x 10^5 V would make 1.4 x 10^7 V/m correct?

That's what I'd get.

Does it make sense to add values of sigma to find the electric field, as was done in the Cramster problem I found online?

I couldn't say. I didn't look at it.
 

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