- #1

Brandone

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## Homework Statement

An electron starts from one plate of a charged closely spaced (vertical) parallel plate arrangement with a velocity of 1.63x10

^{4}m/s to the right. Its speed on reaching the other plate, 2.10 cm away, is 4.15x10

^{4}m/s.

...

If the plates are square with an edge length of 25.4 cm, determine the charge on each.

*Given and Known:*v

_{i}= 1.63x10

^{4}

v

_{f}= 4.15x10

^{4}

d = 2.10 cm

A = (25.4 cm)

^{2}= (0.0645 m

^{2})

*m*

_{e}= 9.109x10

^{-31}kg

*q*

_{e}= 1.602x10

^{-19}C

k = 9.00 x 10

^{9}Nm

^{2}/C

^{2}

## Homework Equations

a = (v

_{f}-v

_{i}) / (t

_{f}-t

_{i})

F =

*m*a

E = F

_{q}/ q

Q = EA / 4[pi]k

## The Attempt at a Solution

**I used:**

*First,*F =

*m*a, and

E = F

_{q}/ q

And got:

E =

*m*a / q

E = (2.52x10

^{4}-N/C) (9.109x10

^{-31}kg) / (1.602x10

^{-19}C)

E = 6.823 x 10

^{-6}N/C

**,**

*Then*Q = EA/4[Pi]k

Q = (6.823 x 10

^{-6}N/C) (0.0645 m

^{2}) / 4[Pi](9.00x10

^{9}Nm

^{2}/C

^{2}

**Q = 3.89 x 10**

^{4}CSo, this does not agree with the answer key's 1.13x10

^{-13}C, but it seems like I'm making sensible steps toward the answer.

Guidance, please?