{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW8Sol - Platt David Homework 8 Due 4:00 am Inst Ken Shih...

This preview shows pages 1–3. Sign up to view the full content.

Platt, David – Homework 8 – Due: Sep 26 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. For 003 you need give your answer in unit of pF (1pF=e-12*F). 001 (part 1 of 2) 10 points Consider a air-filled parallel-plate capacitor with plates of length 8 . 8 cm, width 6 . 78 cm, spaced a distance 1 . 86 cm apart. Now imagine that a dielectric slab with di- electric constant 2 . 7 is inserted a length 4 . 8 cm into the capacitor. The slab has the same width as the plates. The capacitor is com- pletely filled with dielectric material down to a length of 4 . 8 cm. A battery is connected to the plates so that they are at a constant potential 0 . 83 V while the dielectric is inserted. 8 . 8 cm 4 . 8 cm 1 . 86 cm dielectric none - air bottom 1 . 0 dielectric constant top 2 . 7 What is the ratio of the new potential en- ergy to the potential energy before the inser- tion of the dielectric? Edge effects can be neglected. Correct answer: 1 . 92727 . Explanation: Let : w = 6 . 78 cm , d = 1 . 86 cm , L = 8 . 8 cm , = 4 . 8 cm , κ = 2 . 7 , and V = 0 . 83 V . The given capacitor can be considered as two capacitors in parallel, as shown below. C 1 C 2 The potential energy in a capacitor is U = 1 2 C V 2 , so the ratio is U 0 U = C 0 V 0 2 C V 2 . Since the electric potential difference (volt- age) across the capacitor is constant, U 0 U = C 0 C . (1) The capacitance of a parallel plate capaci- tor is C = ² 0 · area separation = ² 0 w L d . (2) After the insertion of the dielectric, the ca- pacitor can be considered in two parts: the part filled with dielectric, and the part with no dielectric. Since the dielectric is inserted with a length ‘ , C 0 = C 0 1 + C 2 , where (3) C 0 1 = ² 0 w κ ‘ d C 2 = ² 0 w ( L - ) d . The total capacitance of the new system is C 0 = C 0 1 + C 2 = ( L - + κ ‘ ) ² 0 w d . (4) Therefore, we have U 0 U = C 0 C = n L + [ κ - 1] o ² 0 w d L ² 0 w d

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Platt, David – Homework 8 – Due: Sep 26 2005, 4:00 am – Inst: Ken Shih 2 = 1 + L h κ - 1 i (5) = 1 + (4 . 8 cm) (8 . 8 cm) h (2 . 7) - 1 i = 1 . 92727 . When the dielectric is fully inserted = L , so U 0 U = 1 + L L h κ - 1 i = κ . (6) 002 (part 2 of 2) 10 points From this, we can deduce that if a person inserts the slab part way into the conductor, and then releases it, the capacitor will exert a force on the dielectric in what direction?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}