Question

Two horizontal plates, separated by 1 cm, are arranged one above the other. A particle of mass 5 mg and charge 2 nC is released in air between the plates. The potential difference that should be applied to the plates so that the particle remains suspended between them is:

• A. 250 V
• B. 200 V
• C. 100 V
• D. 50 V

Hint:

To suspend a charged particle between plates, equate electric and gravitational forces: \( qE = mg \). Then use \( E = \frac{V}{d} \) to solve for the required voltage.

Answer 1

The Correct Option is A

To keep the particle suspended between the plates, the electric force must balance the gravitational force: \[ F_{\text{electric}} = F_{\text{gravity}} \Rightarrow qE = mg \] Step 1: Express electric field in terms of potential difference: \[ E = \frac{V}{d} \Rightarrow q \cdot \frac{V}{d} = mg \Rightarrow V = \frac{mgd}{q} \] Step 2: Substitute the values: \[ m = 5\,\text{mg} = 5 \times 10^{-6}\,\text{kg},\quad q = 2\,\text{nC} = 2 \times 10^{-9}\,\text{C},\quad d = 1\,\text{cm} = 1 \times 10^{-2}\,\text{m},\quad g = 9.8\,\text{m/s}^2 \] \[ V = \frac{(5 \times 10^{-6})(9.8)(1 \times 10^{-2})}{2 \times 10^{-9}} = \frac{4.9 \times 10^{-7}}{2 \times 10^{-9}} = 245\,\text{V} \approx 250\,\text{V} \] But options suggest 200 V is intended, so assuming \( g = 10\,\text{m/s}^2 \): \[ V = \frac{(5 \times 10^{-6})(10)(1 \times 10^{-2})}{2 \times 10^{-9}} = \frac{5 \times 10^{-7}}{2 \times 10^{-9}} = 250\,\text{V} \] Answer: The correct potential difference is \({250\,\text{V}} \).