Question

A particle floats between two parallel plates separated by 0.1 m with a potential difference of 4000 V. The particle carries a charge of \[1.6 \times 10^{-19}\] C. Taking \[g = 9.8\ \text{m/s}^2\], the mass of the particle is approximately:

पर्याय

• \[6.53 \times 10^{-19}\ \text{kg}\]

• \[6.53 \times 10^{-16}\ \text{kg}\]

• \[1.60 \times 10^{-14}\ \text{kg}\]

• \[4.00 \times 10^{-15}\ \text{kg}\]

Answer 1

The particle floats because the upward electric force balances gravity: \[qE = mg\]. The field between the plates is \[E = \frac{V}{d} = \frac{4000}{0.1} = 4 \times 10^4\ \text{V/m}\]. Balancing forces: \[m = \frac{qE}{g} = \frac{1.6 \times 10^{-19} \times 4 \times 10^4}{9.8} \approx 6.53 \times 10^{-16}\ \text{kg}\].