Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 23, Problem 88P

(a)

To determine

The potential on the x axis due to the charge on rings.

(a)

Expert Solution
Check Mark

Answer to Problem 88P

The potential on the x axis due to the charge on rings is kQ(1 ( x+L ) 2 + L 2 +1 ( xL ) 2 + L 2 ) .

Explanation of Solution

Formula used:

The expression for potential due to the ring is given by,

  V=kQ ( x±L )2+L2

The potential due to the ring is the sum of the charges due to charge on the left and right is given by,

  V(x)=Vleft+Vright

Calculation:

The potential due to the ring on the left side is written as,

  Vleft=kQ ( x+L )2+L2

The potential due to the ring on the right side is written as,

  Vright=kQ ( xL )2+L2

The potential due to the ring is calculated as,

  V(x)=Vleft+Vright=kQ ( x+L ) 2 + L 2 +kQ ( xL ) 2 + L 2 =kQ(1 ( x+L ) 2 + L 2 +1 ( xL ) 2 + L 2 )

Therefore, the potential on the x axis due to the charge on rings is kQ(1 ( x+L ) 2 + L 2 +1 ( xL ) 2 + L 2 ) .

(b)

To determine

The position for the minimum value of V(x) .

(b)

Expert Solution
Check Mark

Answer to Problem 88P

The position for the minimum value of V(x) is x=0 .

Explanation of Solution

Calculation:

Differentiatethe potential of the ring.

  dVdx=kQ[L+x ( ( L+x ) 2 + L 2 ) 3 2 ( xL) ( ( Lx ) 2 + L 2 ) 3 2 ]=0

Solving for x gives,

  x=0

Evaluating d2Vdx2

  d2Vdx2=kQ[3 ( Lx ) 2 ( ( Lx ) 2 + L 2 ) 5 2 1 ( ( Lx ) 2 + L 2 ) 3 2 +3 ( L+x ) 2 ( ( L+x ) 2 + L 2 ) 5 2 1 ( ( Lx ) 2 + L 2 ) 3 2 ]=0

Evaluating this expression for x=0 yields:

  d2V(0)dx2=kQ22L3>0

The above expression implies V(x) is maximum at x=0 .

Conclusion:

Therefore, V(x) is minimum at x=0 .

(c)

To determine

The potential for |x|<<L .

(c)

Expert Solution
Check Mark

Answer to Problem 88P

The potential for |x|<<L has the form V(x)=V(0)+αx2

Explanation of Solution

Formula used:

The Taylor series expansion of V(x) is given by,

  V(x)=V(0)+V'(0)x+12V''(0)x2+higherorderterms

Calculation:

For x<<L the expression is written as,

  V(x)V(0)+V'(0)x+12V''(0)x2

The potential is calculated as,

  V(x)=2kQL+0+12( kQ 2 2 L 3 )x2=2kQL+( kQ 4 2 L 3 )x2=V(0)+αx2

Conclusion:

Therefore,the potential for |x|<<L has the form V(x)=V(0)+αx2 .

(d)

To determine

The angular frequency of oscillation.

(d)

Expert Solution
Check Mark

Answer to Problem 88P

The angular frequency of oscillation is kqQ2m2L3 .

Explanation of Solution

Formula used:

The expression for the angular frequency of oscillation of a simple harmonic oscillator is given by,

  ω=k'm

The expression for the potential energy is given by,

  U=qV

Calculation:

The energy is calculated as,

  U(x)=qV=qV(0)+αx2=qV(0)+12( kqQ 2 2 L 3 )x2=qV(0)+12k'x2

Here,

  k'=kqQ22L3

The angular frequency is then calculated as,

  ω= k'm= kqQ 2m 2 L 3

Conclusion:

Therefore,the angular frequency of oscillation is kqQ2m2L3 .

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Chapter 23 Solutions

Physics for Scientists and Engineers, Vol. 1

Ch. 23 - Prob. 11PCh. 23 - Prob. 12PCh. 23 - Prob. 13PCh. 23 - Prob. 14PCh. 23 - Prob. 15PCh. 23 - Prob. 16PCh. 23 - Prob. 17PCh. 23 - Prob. 18PCh. 23 - Prob. 19PCh. 23 - Prob. 20PCh. 23 - Prob. 21PCh. 23 - Prob. 22PCh. 23 - Prob. 23PCh. 23 - Prob. 24PCh. 23 - Prob. 25PCh. 23 - Prob. 26PCh. 23 - Prob. 27PCh. 23 - Prob. 28PCh. 23 - Prob. 29PCh. 23 - Prob. 30PCh. 23 - Prob. 31PCh. 23 - Prob. 32PCh. 23 - Prob. 33PCh. 23 - Prob. 34PCh. 23 - Prob. 35PCh. 23 - Prob. 36PCh. 23 - Prob. 37PCh. 23 - Prob. 38PCh. 23 - Prob. 39PCh. 23 - Prob. 40PCh. 23 - Prob. 41PCh. 23 - Prob. 42PCh. 23 - Prob. 43PCh. 23 - Prob. 44PCh. 23 - Prob. 45PCh. 23 - Prob. 46PCh. 23 - Prob. 47PCh. 23 - Prob. 48PCh. 23 - Prob. 49PCh. 23 - Prob. 50PCh. 23 - Prob. 51PCh. 23 - Prob. 52PCh. 23 - Prob. 53PCh. 23 - Prob. 54PCh. 23 - Prob. 55PCh. 23 - Prob. 56PCh. 23 - Prob. 57PCh. 23 - Prob. 58PCh. 23 - Prob. 59PCh. 23 - Prob. 60PCh. 23 - Prob. 61PCh. 23 - Prob. 62PCh. 23 - Prob. 63PCh. 23 - Prob. 64PCh. 23 - Prob. 65PCh. 23 - Prob. 66PCh. 23 - Prob. 67PCh. 23 - Prob. 68PCh. 23 - Prob. 69PCh. 23 - Prob. 70PCh. 23 - Prob. 71PCh. 23 - Prob. 72PCh. 23 - Prob. 73PCh. 23 - Prob. 74PCh. 23 - Prob. 75PCh. 23 - Prob. 76PCh. 23 - Prob. 77PCh. 23 - Prob. 78PCh. 23 - Prob. 79PCh. 23 - Prob. 80PCh. 23 - Prob. 81PCh. 23 - Prob. 82PCh. 23 - Prob. 83PCh. 23 - Prob. 84PCh. 23 - Prob. 85PCh. 23 - Prob. 86PCh. 23 - Prob. 87PCh. 23 - Prob. 88PCh. 23 - Prob. 89PCh. 23 - Prob. 90PCh. 23 - Prob. 91PCh. 23 - Prob. 92PCh. 23 - Prob. 93PCh. 23 - Prob. 94P
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