Two metal plates have potential differences of 300 V and are 0.01 m apart. A charge particle of mass 1.96 x 10^-15 kg id held in equilibrium between the plates of the capacitor . Then the electric field is
Correct answer: B. 3 x 10^4 V m-1
- A. 3 x 10^2 V m-1
- B. 3 x 10^4 V m-1
- C. 3Vm-1
- D. 3 x 10^-4 V m-1
Explanation
The electric field E between two plates is calculated using the formula E = V/d, where V is the potential difference and d is the distance between the plates. Here, V = 300 V and d = 0.01 m. Substituting these values, we get E = 300 V / 0.01 m = 3 × 104 V m-1. Thus, the correct answer is Option B. Other options are incorrect as they do not use the correct formula or values for the calculation.
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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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