Question

Consider two conducting plates $A$ and $B$ between which the potential difference is$5V$, plate $A$ being at a higher potential. A proton (${p^ + }$or $p$) and an electron (${e^ - }$ or $e$) are released at plates $A$ and $B$ respectively. Two particles then move towards the opposite plates – the proton ${p^ + }$ to plate $B$ and the electron $e$ to plate $A$. Which one will have a larger velocity when they reach their respective destination plates?
a) Both will have the same velocity.
b) The electron will have the larger velocity.
c) The proton will have will have the larger velocity
d) None will be able to reach the destination point.

Answer 1

Hint : Electrons ‘$e$’ are a type of subatomic particle with a negative charge. The proton ‘${p^ + }$’ is a type of subatomic particle which has almost the same charge as an electron but it is positive while an electron is negative. Neutron ${n^0}$ is a type of subatomic particle which has no charge, or is neutral.

Complete step-by-step solution:
We have two conducting plates $A$ and $B$ with a potential difference of $5V$between them , and conducting plate $A$ has higher potential than conducting plate $B$. Positive subatomic particle , aka Proton is released at plate $B$ while negative subatomic particle, aka Electron is released at plate $A$ respectively.
For easy understanding we will assume Proton to be ‘${p^ + }$’ and electron to be ‘$e$’
We know that ‘${p^ + }$’ and ‘$e$’ share almost the same charge, but ‘$e$’ is negative while ‘${p^ + }$’carries positive charge.
Charge of ‘${p^ + }$’ and/or ‘$e$’ = $1.60{\rm{ }} \times {\rm{ }}{10^{ - 19}}\;coulombs$
But even after sharing a similar charge, the mass of proton and electron is not the same. In fact, Electron ‘$e$’ is approximately 2000 times lighter than proton ‘${p^ + }$’
Mass of ‘$e$’ = $9.10938356{\rm{ }} \times {\rm{ }}{10^{ - 31}}\;kilograms$
Which can be rounded up to $\approx 9.1 \times {\rm{ }}{10^{ - 31}}kgs$
Mass of ‘${p^ + }$’ = $1.6726219{\rm{ }} \times {\rm{ }}{10^{ - 27}}\;kilograms$
Which can be rounded off to $\approx 1.67{\rm{ }} \times {\rm{ }}{10^{ - 27}}\;kgs$
It is clear that proton ‘${p^ + }$’ is heavier than electron ‘$e$’
$mass({p^{ + )}} > mass({e^ - })$
And taking reference from Newton’s second law , which states the relationship between acceleration and mass, i.e. mass and acceleration are inversely proportional if force is constant, we can say that the proton ‘${p^ + }$’ will have less acceleration than what the electron ‘$e$’ will have.
$acceleration({p^{ + )}} < acceleration({e^ - })$
Between the two conducting plates given, there is a uniform electric field present and therefore same charge is being applied on both proton ‘${p^ + }$’ and electron ‘$e$’.
But since proton ‘${p^ + }$’ has less acceleration than electron ‘$e$’, it will have lesser velocity achieved than electron ‘$e$’ too.
$\therefore velocity({p^{ + )}} < velocity({e^ - })$
So to conclude, we can say that option (b) is correct.

Note: Capacitance between two conducting plates is a function of the effective plate area, amount of separation between the plates and the dielectric constant of the medium between them which is usually considered air. In a parallel plate capacitor, usually capacitance is directly proportional to surface area and inversely proportional to the separation distance between the plates.