Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.70 cm distant from the first, in a time interval of 1.48×10−6 s .
Find the magnitude of the electric field.


Find the speed of the proton when it strikes the negatively charged plate.

Answer 1

1) Magnitude of the electric field

The distance traveled by the proton is equal to the distance between the two plates:

d = 1.70 cm = 0.017 m

And the proton takes $t=1.48 \cdot 10^{-6}s$ to cover this distance. From these data, we can find the acceleration experienced by the proton

$a=\frac{2d}{t^2}=\frac{2(0.017 m)}{(1.48 \cdot 10^{-6} s)^2}=1.55 \cdot 10^{10} m/s^2$

But we know that the electric force F exerted on the proton is equal to the proton mass (m) times the acceleration (a):

$F=ma=(1.67 \cdot 10^{-27} kg)(1.55 \cdot 10^{10} m/s^2)=2.59 \cdot 10^{-17} N$

and since the electric force is equal to the electric field strength, E, times the proton charge, q: F=qE, we can find E:

$E=\frac{F}{q}=\frac{2.59 \cdot 10^{-17} N}{1.6 \cdot 10^{-19} C}=161.88 N/C$

2) Speed of the proton as it strikes the negative plate

Since the proton starts from rest, its speed at time t is given by:

$v(t) = at$

substituting the values of acceleration and time that we found in the previous part of the problem, we can calculate the speed of the proton:

$v=at=(1.55 \cdot 10^{10} m/s^2)(1.48 \cdot 10^{-6} s)=22940 m/s$

Answer 2

The magnitude of the electric field can be calculated using the equation

$E = \frac{F}{q}$, where F is the Force acting on the proton and q is the charge of the proton.

We know that the charge of the proton is $q = 1.6 x 10^{-19} C$

We have to calculate the Force first.

We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.

We know that the mass of the proton $m = (1.67) X 10^{-27} kg$

So it turns out that we have to calculate acceleration before anything else.

In order to calculate the acceleration, we make use of the following data from the question:

Initial Velocity of the proton $V_{i} = 0$

Distance traveled  $D = 1.70 cm$ = 0.017 m

Time taken for the travel between the plates $t = (1.48) X 10^{-6} s$

Acceleration a = ?

Using the equation, $D = V_{i}t + \frac{1}{2} at^{2}$, we get

Knowing that initial velocity is 0, the equation reduces to $D = \frac{1}{2}at^{2}$

Rearranging the equation so as to make a the subject of the formula, we have

$a = \frac{2D}{t^{2} }$

Plugging in the numbers and simplifying gives us a = 1.5 x $10^{10} m/s^{2}$

We can now calculate the Force using F = ma

Plugging in the known values, we get F = 2.5 x $10^{-17} N$

Using this, we can calculate E through the equation $E = \frac{F}{q}$

Plugging the numbers and simplifying gets us E = 156.25 N/C

Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C

B) To calculate the Final Velocity of the proton, we can make use of the equation

$V_{f} = V_{i} + at$

Plugging the numbers in and simplifying gets us $V_{f} = (2.22) * 10^{4} m/s$