Question 808: Electrostats & Capacitance

Question

A small particle carrying a negative charge of $1.6 	imes 10^{-19} \mathrm{C}$ is suspended in equilibrium between two horizontal metal plates 8 cm apart having a potential difference of 980 V across them. The mass of the particle is $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2ight]$

Accepted Answer

First, find the electric field $$E$$ between the plates using the formula:
$$E = \frac{V}{d},$$
where
$$V = 980\,	ext{V}$$ is the potential difference,
$$d = 8\,	ext{cm} = 0.08\,	ext{m}$$ is the separation between the plates.
Substituting the values:
$$E = \frac{980}{0.08} = 12250\,	ext{V/m}.$$
The charged particle experiences an electric force given by:
$$F_e = qE,$$
where
$$q = 1.6 	imes 10^{-19}\,	ext{C}$$ is the charge of the particle.
Therefore,
$$F_e = \left(1.6 	imes 10^{-19}ight) 	imes 12250 = 1.96 	imes 10^{-15}\,	ext{N}.$$
Since the particle is suspended in equilibrium, the upward electric force balances the downward gravitational force. The gravitational force is:
$$F_g = mg,$$
where
$$g = 9.8\,	ext{m/s}^2$$ is the acceleration due to gravity,
$$m$$ is the mass of the particle.
Setting the forces equal:
$$qE = mg.$$
Solve for the mass $$m$$:
$$m = \frac{qE}{g} = \frac{1.96 	imes 10^{-15}}{9.8} \approx 2 	imes 10^{-16}\,	ext{kg}.$$
Thus, the correct option is Option A:
$$2 	imes 10^{-16}\,	ext{kg}.$$

Answer

$2 	imes 10^{-16} \mathrm{~kg}$

Answer

$2.2 	imes 10^{-16} \mathrm{~kg}$

Answer

$20 	imes 10^{-16} \mathrm{~kg}$

Answer

$4 	imes 10^{-16} \mathrm{~kg}$

Electrostats & Capacitance

A small particle carrying a negative charge of 1.6 \times 10^{-19} \mathrm{C} is suspended in equilibrium between two horizontal metal plates 8 cm apart having a potential difference of 980 V across them. The mass of the particle is \left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right]

Choose the correct answer: