Question

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s.

What will be the final speed of an electron released from rest at the negative plate?

Express your answer to two significant figures and include the appropriate units

V=_________

Answer 1

Concepts and reason

The concept used to solve this problem is speed of electron. The proton is released in a uniform electric field. It is naturally attracted towards negatively charged plate. since the electric field is uniform the motion will be under the influence of a constant force and hence proton will experience uniform acceleration. Similar will be the case with electron. But, motion is towards negative plate. First establish a relation between acceleration of photon and acceleration of electron, using Newton's equation of motion. Then, using knowledge of mechanics, employing equations that describe motion of a particle under uniform acceleration calculate the velocity of electron.

Fundamentals

The force experienced by a charged particle in a uniform electric field is given by, \(F=E q\)

Here, \(F\) is the force on the particle, \(E\) is the uniform electric field and \(q\) is the charge of the particle. Newton's law of motion is represented in equation form as, \(F=m a\)

Here, \(F\) is the force, \(m\) is the mass of the particle and \(a\) is acceleration. For motion with constant acceleration, we can write, \(V^{2}-U^{2}=2 a S\)

Here, \(V\) is final velocity, \(U\) is initial velocity, \(a\) is acceleration and \(S\) is displacement.

The ratio of mass of proton to the mass of electron is 1836.153 .

 

Calculate the Force experienced by the charged particle. The force experienced by a charged particle in a uniform electric field is given by,

$$ F=E q $$

Here, \(F\) is the force on the particle, \(E\) is the uniform electric field and \(q\) is the charge of the particle. since, both electric field and the charge of the particle remains constant for electron and proton we can write,

\(F_{p}=F_{e}\)

Here, \(F_{p}\) stands for force experienced by proton and \(F_{e}\) stands for force experienced by electron.

Substitute for \(F\) using Newton's equation of motion.

\(F_{p}=F_{e}\)

\(m_{p} a_{p}=m_{e} a_{e}\)

Solve for \(a_{p}\)

\(a_{p}=\frac{m_{e} a_{e}}{m_{p}}\)

The force experienced by both proton and electron are same, the acceleration of these particles depends only on their masses. Hence, using equation of motions a relation between acceleration of particles and their masses can be established.

 

Calculate the distance between the plates, which is equal to distance traveled by photon when it's velocity reaches

to \(45000 \mathrm{~m} / \mathrm{s}\)

For motion with constant acceleration we can write,

\(V_{p}^{2}-U_{p}^{2}=2 a_{p} S\)

Here, \(V_{p}\) is final velocity of proton, \(U_{p}\) is initial velocity of proton, \(a_{p}\) is acceleration of proton and \(S\) is displacement.

Substitute for \(45000 \mathrm{~m} / \mathrm{s}\) for \(V_{p}, 0 \mathrm{~m} / \mathrm{s}\) for \(U_{p, \text { and }} a_{p} \mathrm{~m} / \mathrm{s}^{2}\) for \(a\)

\((45000 \mathrm{~m} / \mathrm{s})^{2}-0=2\left(a_{p} \mathrm{~m} / \mathrm{s}^{2}\right) S\)

Solve for \(S\).

\(S=\frac{(45000 \mathrm{~m} / \mathrm{s})^{2}}{2\left(a_{p} \mathrm{~m} / \mathrm{s}^{2}\right)}\)

\(=\frac{(45000)^{2}}{2 a_{p}} \mathrm{~m}\)

Proton is uniformly accelerated, knowing the values of initial velocity, final velocity and acceleration of the particle, we can calculate distance traveled by the photon. since the final velocity is velocity of proton when it reaches negatively charged plate, it is also equal to distance between the plates.

 

We can find final velocity of the electron using equation,

$$ V_{e}^{2}-U_{e}^{2}=2 a_{e} S $$

Here, \(V_{e}\) is final velocity of electron, \(U_{e}\) is initial velocity of electron, \(a_{e}\) is acceleration of electron and \(S\) is displacement.

Substitute \(0 \mathrm{~m} / \mathrm{s}\) for \(U_{e,} a_{p} \mathrm{~m} / \mathrm{s}^{2}\) for \(a\), and \(\frac{(45000)^{2}}{2 a_{p}} \mathrm{~m}\) for \(S\).

\(V_{e}^{2}-0=2\left(a_{e} \mathrm{~m} / \mathrm{s}^{2}\right)\left(\frac{45000^{2}}{2 a_{p}} \mathrm{~m}\right)\)

Simplify for \(V_{e}\)

\(V_{e}=\left[a_{e} \frac{45000^{2}}{a_{p}} \mathrm{~m}^{2} / \mathrm{s}^{2}\right]^{\frac{1}{2}}\)

$$ =\left(\frac{a_{e}}{a_{p}}\right)^{\frac{1}{2}}(45000 \mathrm{~m} / \mathrm{s}) $$

since, \(m_{p} a_{p}=m_{e} a_{e}\)

Solve for ratio of accelerations.

\(\frac{a_{e}}{a_{p}}=\frac{m_{p}}{m_{e}}\)

Substituting in the equation of \(V_{e}\)

\(V_{e}=\left(\frac{m_{p}}{m_{e}}\right)^{\frac{1}{2}}(45000 \mathrm{~m} / \mathrm{s})\)

Substituting for proton to electron mass ratio as 1836.153 for \(\frac{m_{p}}{m_{e}}\)

\(V_{e}=(1836.153)^{\frac{1}{2}}(45000 \mathrm{~m} / \mathrm{s})\)

\(=(42.850)(45000 \mathrm{~m} / \mathrm{s})\)

\(=19,28,250 \mathrm{~m} / \mathrm{s}\)

\(=1.9 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Final speed of the electron is \(1.9 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

since electron is uniformly accelerated, knowing the values of initial velocity, final velocity and acceleration of the particle, we can calculate distance traveled by the electron. Here, since distance travelled by electron is the distance between the plates, therefore it is also equal to distance travelled by electron.