Two metal plates have a potential difference of 300 V and are 0.01 m apart. A charged particle of mass 1.96x 10⁻¹⁵ kg is held in equilibrium between the plates of the capacitor. Then the electric field is:
Correct answer: B. 3×10⁴ V m⁻¹
- A. 3x 10² V m⁻¹
- B. 3×10⁴ V m⁻¹
- C. 3 Vm⁻¹
- D. 3×10⁻⁴ V m⁻¹
Explanation
The electric field (E) between parallel plates is the potential difference (V) divided by the distance (d) between them. E = V/d = 300 V / 0.01 m = 30,000 V/m or 3 × 10⁴ V/m.
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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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