36.

(a) The initial direction of motion is taken to be the

+

x direction (this is also the direction of 

E). We

use v

2

f

− v

2

i

= 2a∆x with v

f

= 0 and a = 

F /m =

−e 

E/m

e

to solve for distance ∆x:

∆x =

−v

2

i

2a

=

−m

e

v

2

i

−2eE

=

−(9.11 × 10

−31

kg)(5.00

× 10

6

m/s)

2

−2(1.60 × 10

−19

C)(1.00

× 10

3

N/C)

= 7.12

× 10

−2

m .

(b) Eq. 2-17 leads to

t =

∆x

v

avg

=

2∆x

v

i

=

2(7.12

× 10

−2

m)

5.00

× 10

6

m/s

= 2.85

× 10

−8

s .

(c) Using ∆v

2

= 2a∆x with the new value of ∆x, we find

∆K

K

i

=

∆(

1
2

m

e

v

2

)

1
2

m

e

v

2

i

=

∆v

2

v

2

i

=

2a∆x

v

2

i

=

−2eE∆x

m

e

v

2

i

=

−2(1.60 × 10

−19

C)(1.00

× 10

3

N/C)(8.00

× 10

−3

m)

(9.11

× 10

−31

kg)(5.00

× 10

6

m/s)

2

=

−11.2% .