Question

A small particle carrying a negative charge of $1.6 \times 10^{-19} C$ 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 [$g = 9.8 m/s^2$]

• A. $2 \times 10^{-16} kg$
• B. $2.2 \times 10^{-16} kg$
• C. $20 \times 10^{-16} kg$
• D. $4 \times 10^{-16} kg$

Answer 1

Answer: (A)

A $2 \times 10^{-16} kg$

Answer 2

Solution:

1. The electric field between the plates is

   $$E = \frac{V}{d} = \frac{980\,\text{V}}{0.08\,\text{m}} = 1.225 \times 10^4\,\text{V/m}.$$

2. For equilibrium, the electric force (which acts upward on the negative charge) must balance the gravitational force:

   $$|q|E = mg.$$

3. Substituting the given values:

   $$m = \frac{|q|E}{g} = \frac{(1.6 \times 10^{-19}\,\text{C})(1.225 \times 10^4\,\text{V/m})}{9.8\,\text{m/s}^2}.$$

4. Calculate the numerator:

   $$1.6 \times 10^{-19} \times 1.225 \times 10^4 = 1.96 \times 10^{-15}\,\text{N}.$$

5. Now,

   $$m = \frac{1.96 \times 10^{-15}\,\text{N}}{9.8\,\text{m/s}^2} = 2.0 \times 10^{-16}\,\text{kg}.$$

Answer: Option A: $2 \times 10^{-16}\,\text{kg}$

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Minimal Explanation:
Equate gravitational force $mg$ to the magnitude of the electric force $qE$. Compute $E = \frac{V}{d}$ and substitute to find $m$.