How many electrons should be removed from a coin of mass 1.6 g, so that it may float in an electric field intensity of 10⁹ N/C which is directed upwards?
Correct answer: A. 9.8 x 10⁷
- A. 9.8 x 10⁷
- B. 9.8 x 10⁵
- C. 9.8 x 10³
- D. 9.8 x 10¹
Explanation
For the coin to float, the electric force (Fₑ = qE) must balance the gravitational force ( F₉= mg). Setting qE = mg, we find the required charge q = mg/E. The number of electrons is then n = q/e, where e is the charge of a single electron.
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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