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.
This question appeared on the KMU Centralized Admission Test 2021 paper, which you can sit online with every answer explained.
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About Capacitors
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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