Question

Calculate the voltage needed to balance an oil drop carrying $10$ electrons between the plates of a capacitor $5\,mm$ apart. The mass of the drop is $3 \times {10^{ - 16}}kg$ and $g = 10\,m{s^{ - 2}}$.

Answer 1

As we know that to balance any object, the upward and downward forces are always equal and in opposite directions. The upward force is $qE$ ; where $q$ is the charge on electrons and $E$ is the electric field. And the downward force is $qE$ . Then we will calculate the electric field with the help of given facts. And atlast we will equate both the forces to find the voltage needed to balance an oil drop between the plates of the capacitor.

Complete step by step answer:
Given that: Distance between the plates of a capacitor, $r = 5\,mm = 5 \times {10^{ - 3}}m$
Mass of the drop, $m = 3 \times {10^{ - 16}}\,kg$
Downward force exerting on the oil drop, $g = 10\,m{s^{ - 2}}$
As we observe the situation:If the oil drop is between the plates of a capacitor, that means there is exerting a downward force, $mg$. And to balance the oil drop between the plates of a capacitor, there is another equal force exerting in upward direction i.e, $qE$ where, $q$ is the charge on electrons and $E$ is the electric field.

So, the both upward and downward force are equals to each other:
$mg = qE$ ………….eq(i)
Now, we will find the electric field of the oil drop by applying its formula in terms of its voltage and the distance between the plates of a capacitor:-
$E = \dfrac{V}{r} = \dfrac{V}{{5 \times {{10}^{ - 3}}}}$
Now, we need to find the charge on the oil drop that is in between the plates of a capacitor:-
$q = ne$
where, $q$ is the charge of the oil drop, $n$ is the number of electron and, $e$ is the charge on an electron, i.e. $1.6 \times {10^{ - 19}}C$.

Now, we will substitute the number of electron and the charge on an electron:
$\Rightarrow q = 10 \times 1.6 \times {10^{ - 19}}C$
So, we will put the value of electric field and the charge of the oil drop in eq(i):-
$mg = qE \\\ \Rightarrow 3 \times {10^{ - 16}} \times 10 = 10 \times 1.6 \times {10^{ - 19}} \times \dfrac{V}{{5 \times {{10}^{ - 3}}}} \\\ \therefore V = 9.47\,volt \\\$
Hence, the voltage needed to balance an oil drop carrying 10 electrons between the plates of a capacitor $5\,mm$ apart is $9.47\,volt$.

Note: There will be no electric field if the charge is uniform at all places, regardless of the electric potential. Thus, “Electric field is the negative space derivative of electric potential” is a common expression for the relationship between electric field and electric potential.