Elecric charge and electric field

[jena]

Hi,

My problem is:

An electron( mass m=9.11 x 10^-31 kg) is accelerated in the uniform field E( E=1.45 x 10^4 N/C) between two parallel charged plates. The separation of the plates is 1.10 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. With what speed does it leave the hole?

Work

m=9.11 x 10^-31 kg
E=1.45 x 10^4 N/C
distance= 1.10 x 10^.2 m
q= -1.6 x 10^-19 C

qE(s) = ½ mv^2 - ½ mv^2

-1.6 x 10^-19 C= ½ (9.0 x 10^-31 kg)(v)^2

solve for v

v=7.14 x10^-7 m/s

Is this the correct way of approaching the problem, or answer?
Please help

Thank You

 

[whozum]

Energy is the correct method, but you made an error in your calculation.

The electron goes through the E field acquiring kinetic energy equal to the loss in potential energy. Its total potential energy is given by $PE = qE\Delta x$, with delta x being the separation of the two plates. Since it starts at rest and experiences no other forces, the final kinetic energy will be equal the initial potential energy.

$$\frac{1}{2}mv^2 = qE\Delta x$$

You seem to have equated the charge with the final kinetic energy on the step before "solve for v".

 

[FluxCapacitator]

A funnier way of writing the potential energy of an electric field:
QED

 

[jena]

Hi,

Whozum said that I should use this equation:
$$\frac{1}{2}mv^2 = qE\Delta x$$

So all I would have to do is plug everything in right?

My work

(1/2)(9.11 x 10 ^31)V^2=(1.6 x 10 ^-19)(1.45 X 10^4)(.11)

V^2=((1.6 x 10 ^-19)(1.45 X 10^4)(.011)
)/((1/2)(9.11 x 10 ^31))

V=7.49 X 10^-25 m/s

But I looked at the back of the book and gives the answer of 7.49 X10^6 m/s :yuck:

Did I forget something?

Please help and Thank You

 

[HiPPiE]

A funnier way of writing the potential energy of an electric field:
QED

Quantum electrodynamics?
?:(

And uhh... check your electron mass again, I'm pretty sure an electron is lighter than the sun. :rofl:

 

[jena]

Hi,

I see what I did wrong.

Work:

(1/2)(9.11 x 10 ^31)V^2=(1.6 x 10 ^-19)(1.45 X 10^4)(.11)

V^2=((1.6 x 10 ^-19)(1.45 X 10^4)(.011)
)/((1/2)(9.11 x 10 ^-31))

V=7.49 X10^6 m/s

Thanks

 

[jena]

Hi,

The second portion of this question asks:

Show that the gravitational force can be ignored.

I found that I can ignore gravitational force as long as the size of qE / m relative to g. As long as qE / m is much larger than g, gravity can be ignored. Gravity is very easy to account for, of course : simply add mg to the free-body diagram and go from there.

So if that's the case couldn't I just find the acceleration of the problem, since I already found the velocity and see if this is true.

Thank You