Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 23, Problem 34P

(a)

To determine

The total electric potential at a large distance from the quadrupole.

(a)

Expert Solution
Check Mark

Answer to Problem 34P

The total electric potential at a large distance from the quadrupole is 2kBcos2θr3

Explanation of Solution

Formula used:

The expression for the electric potential due to quadrupole is given as,

  Vquadrupole=V++V

Here, V is the electric potential at a large distance from dipole.

Calculation:

  Physics for Scientists and Engineers, Chapter 23, Problem 34P

Figure (1)

The electric potential at a large distance from dipole is calculated as,

  V=kqLcosθr2

The electric potential due to quadrupole is calculated as,

  Vquadrupole=V++V=kqLcosθr12kqLcosθr22

  (r2r1=Lcosθr2+r1=2rr1=r2=r)

Substitute the values,

  Vquadrupole=kqLcosθ( ( r 2 2 r 1 2 ) r 1 2 r 2 2 )=kqLcosθ( 2r( Lcosθ ) r 4 )=2kqL2 cos2θr3(B=qL2)=2kB cos2θr3

Conclusion:

Therefore, the total electric potential at a large distance from the quadrupole is 2kBcos2θr3

(b)

To determine

The electric field for Z>>L .

(b)

Expert Solution
Check Mark

Answer to Problem 34P

The electric field for Z>>L is E=(6kB/z4)k^ .

Explanation of Solution

Formula used:

The expression for the electric potential due to quadrupole on the z -axis is given as,

  Vz=2kBcos2θr3

Calculation:

On z -axis the angle θ will be zero and the distance r is equal to z .

  Vz=2kB cos2θr3=2kB cos2(0)z3=2kBz3

The electric field on the positive z -axis is calculated as,

  Ez=z( 2kB z 3 )k^=(2kB)z(z 3)k^=( 6kB z 4 )k^

Conclusion:

Therefore, the electric field for Z>>L is E=(6kB/z4)k^ .

(c)

To determine

The result of part (b) will be obtained by the addition of three point charges.

(c)

Expert Solution
Check Mark

Answer to Problem 34P

The result of part (b) will be obtained by the addition of three point charges and the magnitude of the electric field is E=(6kB/z4)k^ .

Explanation of Solution

Formula used:

The expression for the net electric field on the positive z -axis is given as,

  E=(kq ( zL )2)k^+(k( 2q)z2)k^+(kq( z+ L 2 ))k^

Calculation:

The net electric field on the positive z -axis is calculated as,

  E=( kq ( zL ) 2 )k^+( k( 2q ) z 2 )k^+( kq ( z+ L 2 ))k^=kq[1 ( zL ) 22 z 2+1 ( z+L ) 2]k^=kq[1 z 2 ( 1 L z ) 22 z 2+1 z 2 ( 1+ L z ) 2]k^=kqz2[( 1 L z )22+( 1+ L Z )2]k^ ....... (1)

By using binomial theorem,

  (1 L z)2=1+2Lz+3( L z)2+.............(1+ L z)2=12Lz+3( L z)2+.............

Substitute the binomial equation in equation (1)

  E=kqz2[(1+2 L z+3 ( L z ) 2)2+(12 L z+3 ( L z ) 2)]k^=kqz2[3 L 2 z 2+3 L 2 z 2]k^=( 6kB z 4 )k^

Conclusion:

Therefore, the result of part (b) will be obtained by the addition of three point charges and the magnitude of the electric field is E=(6kB/z4)k^ .

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Chapter 23 Solutions

Physics for Scientists and Engineers

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