Home > 0877 > GR0877 Problem 93

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}.

Found a typo? Comment!

Jump to the problem

1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Categories: 0877 Tags: , , , ,
  1. Michael Humpherys
    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

      Michael Humpherys :

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

      Thanks. Fixed.

  2. Nico
    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.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: