# Homework Help: Capacitors - a paradox?

1. Feb 23, 2012

### 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.

1. The problem statement, all variables and given/known data

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.

3. The attempt at a solution

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!

2. Feb 23, 2012

### Staff: Mentor

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).

3. Feb 23, 2012

### lillybeans

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?

4. Feb 23, 2012

### Staff: Mentor

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.

5. Feb 23, 2012

### lillybeans

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?

6. Feb 23, 2012

Sure.