ism_chapter_24

ism_chapter_24 - 241 Chapter 24 Electrostatic Energy and...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 241 Chapter 24 Electrostatic Energy and Capacitance Conceptual Problems *1 • Determine the Concept The capacitance of a parallel-plate capacitor is a function of the surface area of its plates, the separation of these plates, and the electrical properties of the matter between them. The capacitance is, therefore, independent of the voltage across the capacitor. correct. is ) ( c 2 • Determine the Concept The capacitance of a parallel-plate capacitor is a function of the surface area of its plates, the separation of these plates, and the electrical properties of the matter between them. The capacitance is, therefore, independent of the charge of the capacitor. correct. is ) ( c 3 • Determine the Concept True. The energy density of an electrostatic field is given by 2 2 1 e E u ∈ = . 4 • Picture the Problem The energy stored in the electric field of a parallel-plate capacitor is related to the potential difference across the capacitor by . 2 1 QV U = Relate the potential energy stored in the electric field of the capacitor to the potential difference across the capacitor: QV U 2 1 = . doubles doubling Hence, . to al proportion directly is constant, With U V V U Q *5 •• Picture the Problem The energy stored in a capacitor is given by QV U 2 1 = and the capacitance of a parallel-plate capacitor by . d A C ∈ = We can combine these relationships, using the definition of capacitance and the condition that the potential difference across the capacitor is constant, to express U as a function of d. Express the energy stored in the capacitor: QV U 2 1 = Chapter 24 242 Use the definition of capacitance to express the charge of the capacitor: CV Q = Substitute to obtain: 2 2 1 CV U = Express the capacitance of a parallel-plate capacitor in terms of the separation d of its plates: d A C ∈ = where A is the area of one plate. Substitute to obtain: d AV U 2 2 ∈ = Because d U 1 ∝ , doubling the separation of the plates will reduce the energy stored in the capacitor to 1/2 its previous value: correct. is ) ( d 6 •• Picture the Problem Let V represent the initial potential difference between the plates, U the energy stored in the capacitor initially, d the initial separation of the plates, and V ′ , U ′ , and d ′ these physical quantities when the plate separation has been doubled. We can use QV U 2 1 = to relate the energy stored in the capacitor to the potential difference across it and V = Ed to relate the potential difference to the separation of the plates. Express the energy stored in the capacitor before the doubling of the separation of the plates: QV U 2 1 = Express the energy stored in the capacitor after the doubling of the separation of the plates: QV' U' 2 1 = because the charge on the plates does not change....
View Full Document

This homework help was uploaded on 02/26/2008 for the course PHYSICS 11 taught by Professor Licini during the Spring '07 term at Lehigh University .

Page1 / 92

ism_chapter_24 - 241 Chapter 24 Electrostatic Energy and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online