If both the plate area and the plate separation of a parallel-plate capacitor doubled, the capacitance is
Correct answer: C. Unchanged
- A. Doubled
- B. Halved
- C. Unchanged
- D. Triple
Explanation
The capacitance (C) of a parallel-plate capacitor is given by the formula C = (ε₀ * A) / d, where A is the area of the plates, d is the separation between the plates, and ε₀ is the permittivity of free space. When both the plate area (A) and the plate separation (d) are doubled, the new capacitance becomes C' = (ε₀ * (2A)) / (2d) = (ε₀ * A) / d = C. Thus, the capacitance remains unchanged.Option A is incorrect as it mistakenly assumes that only the area increase affects capacitance without accounting for the separation change. Option B incorrectly assumes that the increase in separation alone would reduce capacitance. Option D miscalculates the relationship entirely by suggesting an increase by a factor of three.
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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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