90. (Fourth problem of Cluster)

(a) At its displaced position, its potential energy (using k = 1/4πε

0

) is

U

i

= k

qQ

d

− x

0

+ k

qQ

d + x

0

=

2kqQd

d

2

− x

2

0

.

And at A, the potential energy is

U

A

= 2

k

qQ

d



.

Setting this difference equal to the kinetic energy of the particle (

1
2

mv

2

) and solving for the speed

yields

v =



2 (U

i

− U

A

)

m

=

2 x

0



k q Q

m d (d

2

− x

2

0

)

.

(b) It is straightforward to consider small x

0

(more precisely, x

0

/d

 1) in the above expression (so

that d

2

− x

2

0

≈ d

2

). The result is

v

≈ 2

x

0

d



k q Q

m d

.

(c) Plugging in the given values (converted to SI units) yields v

≈ 19 m/s.

(d) Using the Pythagorean theorem, we now have

U

i

= 2k

−qQ



d

2

+ x

2

0

.

Therefore, (with U

A

in this part equal to the negative of U

A

in the previous part)

v =



2 (U

i

− U

A

)

m

=

2







k q Q

m

1

d

−

1



d

2

+ x

2

0

.

To simplify, the binomial theorem (Appendix E) is employed:

1



d

2

+ x

2

0

≈

1

d

1

−

1

2

x

2

0

d

2



which leads to

v

≈

x

0

d



2k q Q

m d

.