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This was one of the questions on a practice midterm for my physics class. I was wondering why the answer for part 2 is D instead of C and also, in general, how capacitors behave when there are multiple ones that aren't identical. I said that the right surface of the left plate had charge density of +n, so the left surface of the middle plate would have a charge density of -n since the two surfaces are facing each other. Then, since the middle plate is connected to ground and has a voltage of 0V, the total charge on it would be 0C, so the right surface of it would have to have charge density +n. Then, the left surface of the right conductor would have charge density -n, and charge would go to the ground until the voltage on the right plate equaled -50V. So, because E=n/epsilon, the electric field would be the same for both regions in between the plates. I apologize for making this question so long because I wanted to explain my reasoning. Any help would be appreciated. Thanks!

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I was wondering why the answer for part 2 is D instead of C

This is because the electric field is the negative gradient of voltage $$ \vec E = -\nabla V $$

The first plate is at 100V. The second plate is at 0V. The third plate is at -50V. Therefore the E field is twice as much between plates 1 and 2 then between 2 and 3.

Since Capacitance is only a function of geometry and dielectric permittivity, we can say that the Capacitance between plate 1 and 2 is the same as between 2 and 3. Recall the definition of capacitance: $$ C = \frac{Q}V $$ Since C is constant, and V is doubled, this would imply Q is also doubled. So your statements about the charge distribution are not correct.

Then, since the middle plate is connected to ground and has a voltage of 0V, the total charge on it would be 0C...

I think your confusion comes from the fact that you believe the grounded plate must be electrically neutral. But potential is always relative, and we can call whatever we want "ground" or zero volts.

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  • $\begingroup$ Yes, that is exactly why I was confused, thank you! $\endgroup$ – christina Feb 13 '17 at 15:48

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