chapter16

# Chapter16 - Electrical Energy and Capacitance 59 8 There are eight different combinations that use all three capacitors in the circuit These

This preview shows pages 1–4. Sign up to view the full content.

PROBLEM SOLUTIONS 16.1 (a) Because the electron has a negative charge, it experiences a force in the direction opposite to the ﬁ eld and, when released from rest, will move in the negative x direction. The work done on the electron by the ﬁ eld is WF x q E x xx = ( ) = ( ) =− × ( )( ) ΔΔ 1 60 10 375 3 19 .. C N C 20 10 1 92 10 21 8 × ( ) −− m J . (b) The change in the electric potential energy is the negative of the work done on the particle by the ﬁ eld. Thus, Δ PE W =− = − × 192 10 18 . J continued on next page

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

View Full Document
60 Chapter 16 (c) Since the Coulomb force is a conservative force, conservation of energy gives ∆∆ KE PE += 0, or KE m PE PE fe f =− = v 2 20 , and v f e PE m = ( ) = −− × ( ) × 2 2 1 92 10 911 10 18 31 . . J kg = × 205 10 6 . m s in the direction x 16.2 (a) The change in the electric potential energy is the negative of the work done on the particle by the F eld. Thus, PE W qE x qE y qx xy ( ) + ( ) ( ) 0 5 40 10 . ( ) + ( ) −× ( ) =+ × 62 327 32 0 10 5 65 10 C N C m .. 4 J (b) The change in the electrical potential is the change in electric potential energy per unit charge, or V PE q == × 565 10 10 105 4 . J +5.40 C V 6 16.3 The work done by the agent moving the charge out of the cell is WW P E q V e input field C =− − ( )=+ ( ) 160 10 19 . ( ) 90 10 1 4 10 32 0 J C J . 16.4 PE q V q V V ef i = ( ) ( ) , so q PE VV e fi = = × 192 10 320 10 17 . . J +60.0 J C 19 C 16.5 E V d × 25 000 15 10 17 10 2 6 JC m NC . . 16.6 Since potential difference is work per unit charge V W q = , the work done is WqV = ( ) ( ) + ( ) 3 6 10 12 4 3 10 56 C J C J 16.7 (a) E V d × 600 533 10 113 10 3 5 m . . (b) Fq E × ( ) × ( ) 1 60 10 1 13 10 1 80 10 19 5 14 . CN C N (c) WF s =⋅ ( ) ( ) × cos . θ 180 10 533 290 10 14 3 N m cos . 0 4 38 10 17 ° J
Electrical Energy and Capacitance 61 16.8 (a) Using conservation of energy, ΔΔ KE PE += 0, with KE f = 0 since the particle is “stopped,” we have PE KE m i =− =+ × ( 0 1 2 1 2 911 10 23 1 e kg v . ) × ( ) × 2 85 10 3 70 10 7 2 16 .. ms J The required stopping potential is then Δ Δ V PE q == −× × 370 10 160 10 231 10 16 19 . . . J C 3 231 V kV . (b) Being more massive than electrons, protons traveling at the same initial speed will have more initial kinetic energy and require a greater magnitude stopping potent ial . (c) Since Δ V PE q KE q m q stopping = v 2 2 , the ratio of the stopping potential for a proton to that for an electron having the same initial speed is Δ Δ V V me mm i i p e p e pe = −+ −− v v 2 2 2 2 () 16.9 (a) Use conservation of energy KE PE PE KE PE PE se f i ++ ( ) =++ ( ) or Δ KE PE PE ( ) + ( ) + ( ) = 0 Δ KE = 0 since the block is at rest at both beginning and end. Δ PE kx s ( ) 1 2 0 2 max , where x max is the maximum stretch of the spring. Δ PE W QE x e ( )=− =− ( ) max Thus, 0 1 2 0 2 +− ( ) = kx QE x max max , giving x QE k max CV m 78.0 N m × ( ) × ( ) 2 2 350 10 486 10 64 = × = 436 10 436 2 mc m (b) At equilibrium, Σ FFF k x Q E =+= + = 00 , or eq Therefore, x QE k x eq cm = 1 2 218 max .

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.

## This note was uploaded on 04/23/2010 for the course DS SAD taught by Professor Asd during the Spring '10 term at A.T. Still University.

### Page1 / 27

Chapter16 - Electrical Energy and Capacitance 59 8 There are eight different combinations that use all three capacitors in the circuit These

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online