{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions2

# Solutions2 - Physics 208 Assignment 2 Solutions RT Problem...

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

Physics 208, Assignment 2, Solutions, 2/2/09, RT Problem 1. (a) Positive charges “fall” from high to low potential. But electrons are negative. Hence V 2 > V 1 . _ 1 V 2 V 1 V 2 1 electron charge = -|e| V = 2 V - + (b) The battery holds ∆ V constant (equal to its E . M . F . ). Since ∆ V = - integraltext 2 1 Edx , if the path length from 1 to 2 increases, the electric field decreases more or less proportionally. (If the electric field were uni- form the proportionality would be exact.) (c) For the electron, the change of potential energy -| e | ( V 2 - V 1 ) (which is negative) is compensated by a change of kinetic energy mv 2 / 2 (which is positive); 1 2 mv 2 - | e | V = 0 . When the cathode to anode distance is changed, since ∆ V is un- changed, v is unchanged. Problem 2. See figure. (a) All of the charge + Q resides on the outer surface. The electric field is radial with value E ( r ) = 0 for r < b , 0 for b < r < c , kQ r 2 for c < r . (b) See figure for location of charge. 1

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

View Full Document
b c + + + + + + + + + + + + + + + + + + +Q r 2 ~ 1 kQ 2 c kQ c r 1 ~ r 3c 2c c 0 (b) E(r) r 3c 2c c 0 V(r) (d) b b (c) Evaluate V ( r ) by integrating in from (where V = 0 by definition) to r along the horizontal axis; V ( r ) = - integraldisplay r E · dr = kQ c for r < b , kQ c for b < r < c , kQ r for c < r . (d) See figure. Note that the vanishing of E in a region implies only that V is constant in the region, and nothing about the value of the constant. - - - - - - - - - - - - - - - - - - -Q 2 r 2 ~ 1 r 2 ~ b 3c 2c c 0 b 1 ~ r -kQ c + + + + + + + + + + + + + + + + + + r E(r) r V(r) (e) +2Q 3c 2c c 0 (f) -2Q (e) With point charge - 2 Q placed at the origin the charges on the con- ducting shell rearrange themselves. The electric field in the metal must remain zero. Hence a charge - 2 Q must distribute itself uni- formly on the inner surface of the shell. The total charge on the shell
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

Solutions2 - Physics 208 Assignment 2 Solutions RT Problem...

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

View Full Document
Ask a homework question - tutors are online