Calculate the magnitude of the electric field strength between the plates

**looi76** · May 8th 2008, 02:16 AM

Question:

(a)Two flat parallel metal plates, each of length $12.0 \ cm$ , are separated by a distance of $1.5 \ cm$ , as shown in Fig:

The space between the plates in a vacuum.
The potential difference between the plates is $210V$ . The electric field may be assumed to be uniform in the region between the plates and zero outside this region.
Calculate the magnitude of the electric field strength between the plates.

(b) An electron initially travels parallel to the plates along a line mid-way between the plates, as a shown in Fig. The speed of the electrons is $5.0 \times 10^7 ms^{-1}$ .

For the electron between the plates,

(i) determine the magnitude and direction of its acceleration.
(ii) calculate the time for the electron to travel a horizontal distance equal to the length of the plates.

(c) Use your answer in (b) to determine whether the electron will hit one of the plates or emerge from between the plates.

Attempt:

(a) $E = \frac{V}{d} = \frac{210}{1.5 \times 10^{-2}} = 1.4 \times 10^4 \ Vm^{-1}$

(b) $Q.e = m.a$
$Q = 1.6 \times 10^{-19} \ C$
$m = 9.11 \times 10^{-31} \ Kg$
$e = 1.4 \times 10^4 \ Vm^{-1}$

$a = \frac{Q.e}{m} = \frac{1.6 \times 10^{-19} \times 1.4 \times 10^4}{9.11 \times 10^{-31}}$

$a = 2.5 \times 10^{15} \ ms^{-2}$

Direction is RIGHT because the charge is positive.

Are my answers correct? and I need help in solving part (c)

Thnx in advance.

**Danshader** · May 8th 2008, 02:23 AM

you got them correctly.

to solve part c)
find the time for the electron to just complete the length of the plate part b)ii). using this time find the vertical displacement the electron will travel. If the displacement is larger than the distance between the initial path of the electron to the plate 0.75 cm then it can be concluded that the electron hits the plate before exiting the plates.

**looi76** · May 8th 2008, 02:37 AM

Thanks Danshader

(b)(ii) $Time = \frac{Distance}{Speed}$
$Time = \frac{12 \times 10^{-2} \ m}{5.0 \times 10^7}$
$Time = 2.4 \times 10^{-9}s$

Is this right?
How can I get the vertical component? what is the equation?

**Danshader** · May 8th 2008, 02:46 AM

yup correctly calculated time.

to calculated the vertical displacement use the formula:
$<br /> s = ut + \frac{1}{2} at^2<br />$

you already know that the vertical speed of the electron, u = 0
vertical acceleration,a = $2.5\times10^{25}$
time taken,t = $2.4\times 10^{-9}$

so just substitute this values into the formula and obtain the vertical displacement of the electron. if S >0.75 cm electron will touch the plate else it will pass through the plates

**claudenegm** · April 17th 2013, 02:31 AM

Hey guys,

in b(i),
why did we consider the resultant to be $q.E$ ,
why didn't we say that $q.E-mg=ma$ and so:
$a=\frac{q.E}{m}-g$ ?

I understand that $\frac{q.E}{m}$ will be a really big number compared to $g$ , but I believe it would matter with smaller numbers; it would make more sense to consider it that way, no?

Thanks a lot guys!

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