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## GR0877 Problem 93

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) The capacitance of the parallel-plate capacitor filled with dielectric exceeds the vacuum value by a factor of the dielectric constant: $C = k C_{vac}$. Thus, the electromagnetic energy $\displaystyle U = \frac{1}{2} CV_0^2$ (if $V_0$ is unchanged) is increased by the same factor $k$: $\displaystyle U = \frac{1}{2} k C_{vac}V_0^2 = kU_0$. We are left with (B) and (E), but do not rush to choose the second one. You might think that, when you insert a dielectric between the plates, the electric field between them (that is, inside the dielectric) is reduced by the factor of $k$ in comparison with the “vacuum” value $\displaystyle \frac{V_0}{d}$. But you forgot about the battery! In fact, when the dielectric is inserted, the surface charge density on the plates of the capacitor is changed due to the battery. Let’s apply $\int \mathbf{D} \cdot d\mathbf{a} = Q_{fenc}$ to the gaussian “pillbox” one surface of which is inside the positive plate of the capacitor and the other one is inside the dielectric material ($Q_{fenc}$ here is a free charge enclosed by the gaussian surface; pillbox’s surfaces are parallel to the plate’s surface). Noting that $\mathbf{D} = 0$ inside the metal plate, one has $D A = \sigma A$ $\implies$ $D = \sigma$. The electric field inside the dielectric is $\displaystyle E = \frac{D}{\epsilon} = \frac{D}{\epsilon_0 k}$ $\implies$ $\displaystyle E = \frac{\sigma}{\epsilon_0 k}$. We can find $\sigma$ using the fact that $V_0$ is unchanged during the insertion process. $\displaystyle C_{vac} = \frac{Q_{vac}}{V_0} = \frac{\epsilon_0 S}{d}$, $\displaystyle C = \frac{Q}{V_0} = \frac{\epsilon_0 k S}{d}$, so $\displaystyle \sigma = \frac{Q}{S} = k \frac{Q_{vac}}{S} = k \sigma_{vac}$. Finally, $\displaystyle E = \frac{\sigma}{\epsilon_0 k} = \frac{\sigma_{vac}}{\epsilon_0} \equiv \frac{V_0}{d}$.

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Categories: 0877 Tags: , , , ,
1. 07.11.2012 at 21:13

“…charge density on the lates of the capacitor is changed due to the battery.”
“lates” should be “plates”

• 11.11.2012 at 21:12

“…charge density on the lates of the capacitor is changed due to the battery.”
“lates” should be “plates”

Thanks. Fixed.

2. 30.03.2014 at 04:19

Shorter argument for the choice beween E and B: Since the electric field is constant (by symmetry) we have that $latex E = V/d$ , where d is the distance between the capacitor plates. but the voltage and distance d haven’t changed so the field remains unchanged.