Capacitor Paradox: How Can Voltage Stay Constant When Charge Density Increases?

[lillybeans]

So as I was doing my sample physics midterm, I realized a paradox or a concept that I just haven't fully grasped, hence I see a problem.

Homework Statement

Suppose we have two parallel plates of a capacitor with:
capacitance C,
area A,
distance of separation D,
a constant voltage V=10V supplied by the battery that is connected to the capacitor,
electric field E=V/d,
surface charge density σ=(E*ε0)

As we insert a piece of dielectric wth κ=1275 WHILE THE CAPACITOR REMAINS CONNECTED TO THE 10V battery, find the NEW:
Electric field (E1),
charge density σ1,
Capacitance (C1),
Potential Energy U.

The Attempt at a Solution

So here is the paradox.

1. Since capacitor is still connected to the battery, V across capacitor must remain constant, so V1=V=10V.
2. If V is constant, and C increases due to the insertion of the dielectric, then total Q must increase.
3. if total Q increases, then the charge density must increase since A is the same. if σ (charge density) increases, then the new electric field must also increase, since E1=σ1/ε
4. If the electric field increases, the new voltage is given by V1=E1D. D remains the same, E1 increases, so the new voltage increases. However, we were just told that the voltage is held constant at 10V, so how can it increase and stay at 10V at the same time? This does not make sense to me!

Please explain! Many thanks.

 

[gneill]

The field strength in the dielectric is LESS than the field strength for vacuum (or air).

The capacitance goes up (k*C) and the field strength goes down (E/k).

 

[lillybeans]

The field strength in the dielectric is LESS than the field strength for vacuum (or air).

The capacitance goes up (k*C) and the field strength goes down (E/k).

But if the electric field goes down, so must the voltage (V=Ed). However, it is told in the question that the voltage must remain constant. How does that work?

 

[gneill]

But if the electric field goes down, so must the voltage (V=Ed). However, it is told in the question that the voltage must remain constant. How does that work?

Ah. In the case where the voltage is forced to remain constant, E = V/d remains constant. With a dielectric with dielectric constant $\kappa$, the capacitance is given by:

$C = \kappa \epsilon_o \frac{A}{d}$

The charge on the capacitor is then

$Q = V \;C$

so that $Q = V\left(\kappa \epsilon_o \frac{A}{d} \right)$

That makes the charge density

$\sigma = \frac{Q}{A} = V \frac{\kappa \epsilon_o}{d}$

and the field strength

$E = \frac{V}{d} = \frac{\sigma}{\kappa \epsilon_o}$

So the electric field strength is divided by $\kappa$ for the given charge density.

 

[lillybeans]

Ah. In the case where the voltage is forced to remain constant, E = V/d remains constant. With a dielectric with dielectric constant $\kappa$, the capacitance is given by:

$C = \kappa \epsilon_o \frac{A}{d}$

The charge on the capacitor is then

$Q = V \;C$

so that $Q = V\left(\kappa \epsilon_o \frac{A}{d} \right)$

That makes the charge density

$\sigma = \frac{Q}{A} = V \frac{\kappa \epsilon_o}{d}$

and the field strength

$E = \frac{V}{d} = \frac{\sigma}{\kappa \epsilon_o}$

So the electric field strength is divided by $\kappa$ for the given charge density.

Thank you. So is it correct for me to say that after the dielectric has been inserted with the battery connected, V stays the same, E stays the same as before, but σ has gone up?

 

[gneill]

Thank you. So is it correct for me to say that after the dielectric has been inserted with the battery connected, V stays the same, E stays the same as before, but σ has gone up?

Sure.