Question

Two insulating plates are both uniformly charged in such a way that the potential difference between them is ${V_2} - {V_1} = 20V$. (i.e., the plate $2$ is at a higher potential). The plates are separated by $d = 0.1\,m$ and can be treated as infinitely large. An electron is released from rest on the inner surface of the plate $1$. What is its speed when it hits the plate $2$?
($e = 1.6 \times {10^{ - 19}}C$, ${m_e} = 9.11 \times {10^{ - 31}}kg$)
(A) $2.65 \times {10^6}m/s$
(B) $7.02 \times {10^{12}}m/s$
(C) $1.87 \times {10^6}m/s$
(D) $32 \times {10^{ - 19}}m/s$

Answer 1

Here we will use the basic principle of conservation of energy where the kinetic energy is equal to the electrostatic potential energy, which means that the amount of energy acquired by the electron in the electric field is equal to the kinetic energy acquired by the electron.
Formula used:
Kinetic Energy=$\dfrac{1}{2}m{v^2} = qV$=Electrostatic potential energy

Complete step by step answer:
In the problem above, an electron is released from one plate of the insulating surface to another. As it travels, the constant voltage of $20V$ acts on the electron. This constant electric field forces the electron to accelerate so that it keeps gaining momentum and attains a velocity which we need to find. Thus work done on the electron or energy gained by the electron is equal to $qV$, where $V$ is the voltage through which the electron passes and $q$ is the charge on the electron. This is equal to $qV = 1.6 \times {10^{ - 19}}C \times 20V = 32 \times {10^{ - 19}}J$.
Now this energy which is the electrostatic potential energy acquired by the electron in the electric field between the charged region in the two plates is converted into the kinetic energy of the electron.
Thus using the kinetic energy is equal to the change in electrostatic potential energy, we have
$qV = \dfrac{1}{2}m{v^2}$
$\Rightarrow 32 \times {10^{ - 19}}J = \dfrac{1}{2}m{v^2}$.
Substituting the mass of the electron into the equation alongside, we get
$32 \times {10^{ - 19}}J = \dfrac{1}{2}(9.11 \times {10^{ - 31}}kg){v^2}$,
$\Rightarrow {v^2} = \dfrac{{64}}{{9.11}} \times {10^{12}} = 7.0252 \times {10^{12}}$
$\Rightarrow v = 2.65 \times {10^6}$

Therefore the correct answer is option (A).

Note: The mass of the electron is the rest mass as given in the question, although the speed is considerably high; since not very high accuracy is required for the problem above. Also, the voltage through which the electron accelerated is supposed to remain constant with some alterations near the plates.