Electric Field, Potential Between 3 Charged Electrodes

[Bogus_Roads]

Homework Statement

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

 

[gneill]

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.

 

[dimpledur]

Never mind!

 

[Bogus_Roads]

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?

 

[Bogus_Roads]

Maybe they used millimeters instead of centimeters for the distance between the plates.

 

[gneill]

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.