Capacitance of a parallel plate capacitor in the presence of insulator having dielectric constant E is C if the dielectric is removed from plates, then new capacitance will be:
Correct answer: D. C/E
- A. C - E
- B. E.C
- C. C + E
- D. C/E
Explanation
The capacitance of a parallel plate capacitor with a dielectric is given by the formula C = E * (ε₀ * A / d), where E is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between them. When the dielectric is removed, the new capacitance reduces to C/E, as the dielectric constant no longer contributes to increasing the capacitance. Thus, removing the dielectric reduces the capacitance by a factor of the dielectric constant. The other options are incorrect because they either incorrectly suggest an increase or do not accurately reflect the relationship between capacitance and the dielectric constant.
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Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
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