Question

A charged particle of mass $4 \times {10^{ - 13}}\,{\text{kg}}$ is hanging stationary between two horizontal charged plates separated by $2\,{\text{cm}}$ and potential difference between plates is $3.2 \times {10^4}\,{\text{V}}$. Find the charge of the particle.
A. $2.8 \times {10^{ - 15}}\,{\text{C}}$
B. $2.4 \times {10^{ - 20}}\,{\text{C}}$
C. $2.4 \times {10^{ - 15}}\,{\text{C}}$
D. $2.5 \times {10^{ - 18}}\,{\text{C}}$

Answer 1

Use the formula for the electrostatic force in terms of the electric field and charge. Also use the formula for the electric field in terms of the potential difference and separation between the plates. Using the expression for Newton’s second law of motion, write the balanced equation for the weight of the charge and electrostatic force and solve it to calculate the charge on the particle.

Formulae used:
The electrostatic force $F$ is given by
$F = qE$ …… (1)
Here, $q$ is the charge and $E$ is the electric field.
The electric field $E$ between two plates is given by
$E = \dfrac{V}{d}$ …… (2)
Here, $V$ is the potential difference between the two plates and $d$ is the separation between two plates.

Complete step by step answer:
We have given that the mass of the charged particle is $4 \times {10^{ - 13}}\,{\text{kg}}$.
$m = 4 \times {10^{ - 13}}\,{\text{kg}}$
The separation between the two charged plates is $2\,{\text{cm}}$ and the potential difference between the two plates is $3.2 \times {10^4}\,{\text{V}}$.
$d = 2\,{\text{cm}}$
$\Rightarrow V = 3.2 \times {10^4}\,{\text{V}}$
We have asked to calculate the charge on the particle. Let us first convert the unit of separation between the charges plated to the SI system of units.
$d = \left( {2\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)$
$\Rightarrow d = 0.02\,{\text{m}}$

As the charge is stationary between the two plates, the weight of the charge is balanced by the electrostatic force of repulsion between the charge and one of the plates.
$mg = F$
Substitute $qE$ for $F$ in the above equation.
$mg = qE$
Substitute $\dfrac{V}{d}$ for $E$ in the above equation.
$mg = q\dfrac{V}{d}$
$\Rightarrow q = \dfrac{{mgd}}{V}$

Substitute $4 \times {10^{ - 13}}\,{\text{kg}}$ for $m$, $10\,{\text{m/}}{{\text{s}}^2}$ for $g$, $0.02\,{\text{m}}$ for $d$ and $3.2 \times {10^4}\,{\text{V}}$ for $V$ in the above equation.
$\Rightarrow q = \dfrac{{\left( {4 \times {{10}^{ - 13}}\,{\text{kg}}} \right)\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)\left( {0.02\,{\text{m}}} \right)}}{{3.2 \times {{10}^4}\,{\text{V}}}}$
$\therefore q = 2.5 \times {10^{ - 18}}\,{\text{C}}$
Therefore, the charge on the particle is $2.5 \times {10^{ - 18}}\,{\text{C}}$.

Hence, the correct option is D.

Note: The students may think that the weight of the stationary charge must be greater and not equal to the electrostatic force between the plates and charged particle. But since the particle is stationary between the charged plates, these two forces must have the same magnitude and act in the opposite directions.